Differential Equations of First Order and First Degree Exact Differential Equations
1. Consider xy = c. Differentiating w.r.t. \(x: x \frac{d y}{d x}+y=0\)
2. Consider x/y=c. Differentiating w.r.t. \(x: x \frac{d y}{d x}-y=0\)
3. Consider \(e^x \cos y=c\). Differentiating w.r.t. \(x: e^x \cos y-e^x(\sin y) \frac{d y}{d x}=0\)
In the above examples, the differential equations are formed just by the process of differentiation. Such differential equations are called exact equations.
Now from example 1, we have \(x \frac{d y}{d x}+y=0 \Rightarrow x d y+y d x=0 \Rightarrow d(x, y)=0\)
∴ Its solution is \(x y=c_1\)
Similarly, differential equations in examples 2 and 3 can be written in the form \(d\left(\frac{x}{y}\right)=0, d\left(e^x \cos y\right)=0\) and their solutions are respectively \(\frac{x}{y}=c_2\) and \(e^x \cos y=c_3\)
Exact Differential Equations First Order And First Degree Examples
Thus, if a differential equation is expressed as d[f(x,y)] = 0, then its solution is f(x,y) = c.
Note: An exact differential equation can always be derived from its general solution directly by differentiating without any subsequent multiplication, or elimination.
First-Order And First-Degree Differential Equations Examples
Differential Equations of First Order and First Degree Definition
Let \(M(x, y) d x+N(x, y) d y=0\) be a first-order and first-degree differential equation where M, N are real-valued functions defined for some real x,y on some rectangle \(R:\left|x-x_0\right| \leq a ;\left|y-y_0\right| \leq b\).
Then the equation Mdx + Ndy = 0 is said to be an exact differential equation if there exists a function f(x,y) having continuous first partial derivatives in R such that
d\([f(x, y)]=M(x, y) d x+N(x, y) d y \Rightarrow \frac{\partial f}{\partial x} d x+\frac{\partial f}{\partial y} d y=M d x+N d y\)
e.g. 1. \(2 x y d x+x^2 d y=0\) is an exact equation.
For, there exists a function \(x^2 y\) such that
d\(\left({x}^2 y\right)=\frac{\partial}{\partial x}\left(x^2 y\right) d x+\frac{\partial}{\partial y}\left(x^2 y\right) d y \Rightarrow d\left(x^2 y\right)=2 x y d x+x^2 d y\)
e.g. 2. x dx + y dy = 0 is an exact equation.
For, there exists a function \(\frac{x^2+y^2}{2}\) such that
d\(\left(\frac{x^2+y^2}{2}\right)=\frac{\partial}{\partial x}\left(\frac{x^2+y^2}{2}\right) d x+\frac{\partial}{\partial y}\left(\frac{x^2+y^2}{2}\right) d y \Rightarrow d\left(\frac{x^2+y^2}{2}\right)=x d x+y d y\)
Definition And Solutions Of Exact Differential Equations
Theorem 1: If M(x, y), N(x, y) are two real-valued functions that have continuous first partial derivatives on some rectangle \(\mathrm{R}:\left|x-x_0\right| \leq a,\left|y-y_0\right| \leq b\), then a necessary and sufficient condition for the differential equation Mdx+Ndy=0 to be exact, is \(\frac{\partial \mathbf{M}}{\partial y}=\frac{\partial \mathbf{N}}{\partial x} \text { in } R\)
Proof. (1) The condition is necessary.
Let the equation \(\mathrm{M} d x+\mathrm{N} d y=0\) be an exact equation. Now we have to show that \(\frac{\partial \mathrm{M}}{\partial y}=\frac{\partial \mathrm{N}}{\partial x}\)
By definition, there must exist a function f(x, y) having continuous first and second partial derivatives such that \(d[f(x, y)]=\mathrm{M} d x+\mathrm{N} d y\) ……………………..(1)
From total differentiation, we have \(d[f(x, y)]=\frac{\partial f}{\partial x} d x+\frac{\partial f}{\partial y} d y\) ………………….(2)
From (1) and (2): \(\mathrm{M} d x+\mathrm{N} d y=\frac{\partial f}{\partial x} d x+\frac{\partial f}{\partial y} d y\)
so that \(\mathrm{M}=\frac{\partial f}{\partial x}\) ……………….. (3) and \(\mathrm{N}=\frac{\partial f}{\partial y}\) ………………….(4)
Differentiating partially (3) and (4) w.r.t. y and x respectively, we get
\(\frac{\partial \mathrm{M}}{\partial y}=\frac{\partial^2 f}{\partial y \partial x}\) ……………………(5) and \(\frac{\partial \mathrm{N}}{\partial x}=\frac{\partial^2 f}{\partial x \partial y}\) …………………….(6)
for f(x, y) we have \(\frac{\partial^2 f}{\partial y \partial x}=\frac{\partial^2 f}{\partial x \partial y}\)
∴ From (5) and (6): \(\frac{\partial \mathrm{M}}{\partial y}=\frac{\partial \mathrm{N}}{\partial x}\)
Thus if \(\mathrm{M} d x+\mathrm{N} d y=0\) is exact, M and N satisfy the equation \(\frac{\partial \mathrm{M}}{\partial y}=\frac{\partial \mathrm{N}}{\partial x}\)
(2) The condition is sufficient.
Let us now show that if the equation \(\frac{\partial \mathrm{M}}{\partial y}=\frac{\partial \mathrm{N}}{\partial x}\) is satisfied, then Mdx + Ndy = 0 is an exact equation.
For this we have to find a function f(x,y) such that \(d[f(x, y)]=\mathrm{M} d x+\mathrm{N} d y\)
Let us define u(x,y)\(=\int^x \mathrm{M} d x\) ……………….(7)
where \(\int^x\) denotes that while integrating, y is to be treated as constant.
From (7), we have \(\frac{\partial}{\partial x}\left(\int^x \mathrm{M} d x\right)=\frac{\partial u}{\partial x} \Rightarrow \mathrm{M}=\frac{\partial u}{\partial x}\) ……………………(8)
⇒ \(\frac{\partial \mathrm{M}}{\partial y}=\frac{\partial^2 u}{\partial y \partial x}\)
Since => \(\frac{\partial \mathrm{M}}{\partial y}=\frac{\partial \mathrm{N}}{\partial x}\) (given) and \(\frac{\partial^2 u}{\partial y \partial x}=\frac{\partial^2 u}{\partial x \partial y}\)
we have \(\frac{\partial \mathrm{N}}{\partial x}=\frac{\partial^2 u}{\partial x \partial y}=\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial y}\right)\)
Integrating this w.r.t. x treating y as constant, we get \(\mathrm{N}=\frac{\partial u}{\partial y}+\mathrm{C}=\frac{\partial u}{\partial y}+\phi(y)\) ……………………(9)
(∵ the arbitrary constant C may be any function of y)
From (8) and (9) :
⇒ \(\mathrm{M} d x+\mathrm{N} d y=\frac{\partial u}{\partial x} d x+\left[\frac{\partial u}{\partial y}+\phi(y)\right] d y\) = \(\left(\frac{\partial u}{\partial x} d x+\frac{\partial u}{\partial y} d y\right)+\phi(y) d y=d u+\phi(y) d y=d\left[u+\int \phi(y) d y\right]\) ………………….(10) which is an exact differential.
∴ Mdx + Ndy = 0 is an exact differential equation.
Differential Equations of First Order and First Degree Working Rule For Solving An Exact Differential Equation :
1. Compare the given equation with Mdx+Ndy = 0 and find out M and N. Then find out \(\frac{\partial \mathrm{M}}{\partial y} \text { and } \frac{\partial \mathrm{N}}{\partial x}\)
2. Given equation will be exact if \(\frac{\partial \mathrm{M}}{\partial y}=\frac{\partial \mathrm{N}}{\partial x}\)
3. Integrate M partially with respect to x, treating y as constant. Denote this by \(\int^x \mathrm{M} d x\)
4. Integrate only those terms of N, which do not contain x, with respect to y.
5. Equate the sum of the results obtained from (3) and (4) to a constant to obtain the required solution.
∴ The general solution of the given exact differential equation is \(\int^x \mathrm{M} d x+\int\) (terms of N not involving x) dy = c
Differential Equations of First Order and First Degree Equations Reducible To Exact Form
Definition. Let M(x,y)dx + N(x,y)dy = 0 be not an exact differential equation. If M dx + N dy = 0 can be made exact by multiplying it with a suitable function μ (x,y) ≠ 0, , then μ (x,y) is called an integrating factor of M dx + N dy = 0.
Example. Let y dx – x dy = 0 ……………………….. (1) where M = y,N = -x
Then \(\frac{\partial \mathrm{M}}{\partial y}=1, \frac{\partial \mathrm{N}}{\partial x}=-1 \Rightarrow \frac{\partial \mathrm{M}}{\partial y} \neq \frac{\partial \mathrm{N}}{\partial x}\) … (1) is not an exact equation.
Multiplying (1) with \(1 / x^2\), we get: \(\left(y / x^2\right) d x-(1 / x) d y=0\) …………………………(2)
where \(\mathrm{M}=y / x^2, \mathrm{~N}=-1 / x\)
Since \(\frac{\partial \mathrm{M}}{\partial y}=1 / x^2=\frac{\partial \mathrm{N}}{\partial x}\), (2) is an exact equation.
Hence \(1 / x^2\) is an integrating factor of y dx -x dy=0.
Solved Problems On First-Order Exact Differential Equations
Since d(x / y) = \(\frac{y d x-x d y}{y^2}, d[\log (x / y)]=\frac{y d x-x d y}{x y}\) and \(d\left[\text{Tan}^{-1}(x / y)\right]=\frac{y d x-x d y}{x^2+y^2}\), the functions \(1 / y^2, 1 / x y, 1 /\left(x^2+y^2\right)\) are also integrating factors of y dx-x d y=0
Note: From the above example we observe that a differential equation can have more than one integrating factor.
Differential Equations of First Order and First Degree Methods To Find Integrating Factors Of Mdx + Ndy = 0
Method 1. By Inspection: An integrating factor (I.F) of the given equation Mdx + Ndy = 0 can be found by inspection as explained below.
By rearranging the terms of the given equation or (and) by dividing with a suitable function of x and y, the equation thus obtained will contain several parts integrable easily. In this connection, the following exact differentials will be found useful:
(1) \(d(x y)=x d y+y d x\)
(2) \(d[\log (x y)]=\frac{x d y+y d x}{x y}\)
(3) \(d\left(\frac{x}{y}\right)=\frac{y d x-x d y}{y^2}\)
(4) \(d\left(\frac{y}{x}\right)=\frac{x d y-y d x}{x^2}\)
(5) \(d\left(\frac{y^2}{x}\right)=\frac{2 x y d y-y^2 d x}{x^2}\)
(6) \(d\left(\frac{x^2}{y}\right)=\frac{2 x y^{\prime} d x-y^2 d y}{y^2}\)
(7) \(d\left(\frac{y^2}{x^2}\right)=\frac{2 x^2 y d y-2 x y^2 d x}{x^4}\)
(8) \(d\left(\frac{x^2}{y^2}\right)=\frac{2 y^2 x d x-2 y x^2 d y}{y^4}\)
(9) \(d\left(\ Tan^-1 \frac{y}{x}\right)=\frac{x d y-y d x}{x^2+y^2}\)
(10) \(d\left(\ Tan^-1 \frac{x}{y}\right)=\frac{y d x-x d y}{x^2+y^2}\)
(11) \(d\left(\frac{e^x}{y}\right)=\frac{y e^x d x-e^x d y}{y^2}\)
(12) \(d\left(\frac{e^y}{x}\right)=\frac{x e^y d y-e^y d x}{x^2}\)
(13) \(d\left[\log \left(\frac{x}{y}\right)\right]=\frac{y d x-x d y}{x y}\)
(14) \(d\left[\log \left(\frac{y}{x}\right)\right]=\frac{x d y-y d x}{x y}\)
Differential Equations of First Order and First Degree Method 2 To Find An Integrating Factor Of Mdx + Ndy=0
Theorem 2: M(x,y)dx + N(xyy)dy – 0 is a homogeneous differential equation and Mx + Ny ≠ 0, then \(\frac{1}{\mathrm{M} x+\mathrm{N} y}\) is an integrating factor of Mdx+ Ndy = 0.
Proof. Given Mdx+Ndy = 0 is a homogeneous differential equation.
∴ \(\mathrm{M}=x^k \cdot f(y / x), \mathrm{N}=x^k g(y / x)\)
By Euler’s theorem on homogeneous functions, we have \(x \frac{\partial \mathrm{M}}{\partial x}+y \frac{\partial \mathrm{M}}{\partial y}=k \mathrm{M}\) …………………..(2)
and \(x \frac{\partial \mathrm{N}}{\partial x}+y \frac{\partial \mathrm{N}}{\partial y}=k \mathrm{~N}\) ………………………(3)
Let us take \(\mathrm{M}_1=\frac{\mathrm{M}}{\mathrm{Mx}+\mathrm{Ny}}, \mathrm{N}_1=\frac{\mathrm{N}}{\mathrm{Mx}+\mathrm{Ny}} . \text { Then } \frac{\partial \mathrm{M}_1}{\partial y}=\frac{\partial}{\partial y}\left(\frac{\mathrm{M}}{\mathrm{M} x+\mathrm{Ny}}\right)\)
= \(\frac{(\mathrm{M} x+\mathrm{N} y) \frac{\partial \mathrm{M}}{\partial y}-\mathrm{M}\left(x \frac{\partial \mathrm{M}}{\partial y}+\mathrm{N}+y \frac{\partial \mathrm{N}}{\partial y}\right)}{(\mathrm{M} x+\mathrm{N} y)^2}\)
= \(\frac{\mathrm{Ny} y \frac{\partial \mathrm{M}}{\partial y}-\mathrm{MN}-\mathrm{My} \frac{\partial \mathrm{N}}{\partial y}}{(\mathrm{M} x+\mathrm{N} y)^2}\)
⇒ \(\frac{\partial \mathrm{N}_1}{\partial x}=\frac{\partial}{\partial x}\left(\frac{\mathrm{N}}{\mathrm{M} x+\mathrm{N} y}\right)=\frac{(\mathrm{M} x+\mathrm{N} y) \frac{\partial \mathrm{N}}{\partial x}-\mathrm{N}\left(\mathrm{M}+x \frac{\partial \mathrm{M}}{\partial x}+y \frac{\partial \mathrm{N}}{\partial x}\right)}{(\mathrm{M} x+\mathrm{N} y)^2}=\frac{\mathrm{M} x \frac{\partial \mathrm{N}}{\partial x}-\mathrm{MN}-\mathrm{N} x \frac{\partial \mathrm{M}}{\partial x}}{(\mathrm{M} x+\mathrm{N} y)^2}\)
⇒ \(\frac{\partial \mathrm{M}_1}{\partial y}-\frac{\partial \mathrm{N}_1}{\partial x}=\frac{\mathrm{N} y \frac{\partial \mathrm{M}}{\partial y}-\mathrm{MN}-\mathrm{M} y \frac{\partial \mathrm{N}}{\partial y}-\mathrm{M} x \frac{\partial \mathrm{N}}{\partial x}+\mathrm{MN}+\mathrm{N} x \frac{\partial \mathrm{M}}{\partial x}}{(\mathrm{M} x+\mathrm{N} y)^2}\)
= \(\frac{\mathrm{N}\left(y \frac{\partial \mathrm{M}}{\partial y}+x \frac{\partial \mathrm{M}}{\partial x}\right)-\mathrm{M}\left(x \frac{\partial \mathrm{N}}{\partial x}+y \frac{\partial \mathrm{N}}{\partial y}\right)}{(\mathrm{M} x+\mathrm{N} y)^2}=\frac{\mathrm{N}(k \mathrm{M})-\mathrm{M}(k \mathrm{~N})}{(\mathrm{M} x+\mathrm{N} y)^2}=0\)
∴ \(\frac{\partial \mathrm{M}_1}{\partial y}=\frac{\partial \mathrm{N}_1}{\partial x} \Rightarrow \mathrm{M}_1 d x+\mathrm{N}_1 d y=0\) is an exact equation.
⇒ \(\frac{1}{\mathrm{M} x+\mathrm{N} y}(\mathrm{M} d x+\mathrm{N} d y)=0\) is an exact equation.
Hence \(\frac{1}{M x+N y}\) is an integrating factor of Mdx + Ndy = 0.
Note. If Mx + Ny = 0; then, M/N =-y/x. Then the equation Mdx+Ndy = 0 reduces to y dx – x dy = 0. and its solution is x/y = c.
Differential Equations of First Order and First Degree Working Rule In solve Mdx + Ndy = 0
1. The general equation is Mdx + Ndy = 0 …. (1). Observe \(\frac{\partial \mathrm{M}}{\partial y} \neq \frac{\partial \mathrm{N}}{\partial x} \Rightarrow\) (1) is not exact.
2. Observe M and N are homogeneous functions of the same order.
3. Find Mx + Ny and observe it ≠ as 0. Then \(\frac{1}{\mathrm{M} x+\mathrm{N} y}\) is an I.F of(1).
4. Multiply (1) with I.F. to transform it into an exact equation of (1) \(\mathrm{M}_1 d x+\mathrm{N}_1 d y=0\) …………………. (2)
5. Solve (2) to get the general solution of (1)
Differential Equations of First Order and First Degree Method 3 To Find An Integrating Factor Of Mdx+Ndy = 0
Theorem 3: If the equation Mdx+Ndy = 0 is of the form y f(xy) dx + x g (xy) dy = 0 and Mx – Ny ≠ 0, \(\frac{1}{\mathrm{M} x-\mathrm{Ny}}\)then is an integrating factor of Mdx+Ndy=0
Proof. Given equation is Mdx+Ndy = 0 …………………….(1)
Comparing (1) with \(y f(x y) d x+x g(x y) d y=0, \text { we have } \mathrm{M}=y f(x y), \mathrm{N}=x g(x y)\)
Let \(x y=u \Rightarrow \frac{\partial u}{\partial x}=y, \frac{\partial u}{\partial y}=x\) ……………………(2)
Let us take \(\mathrm{M}_1=\frac{\mathrm{M}}{\mathrm{Mx}-\mathrm{Ny}}, \mathrm{N}_1=\frac{\mathrm{N}}{\mathrm{M} x-\mathrm{N} y}\)
⇒ \(\mathrm{M}_1=\frac{y f(x y)}{x y f(x y)-x y g(x y)}=\frac{f}{x(f-g)}, \quad \mathrm{N}_1=\frac{x g(x y)}{x y f(x y)-x y g(x y)}=\frac{g}{y(f-g)}\)
Let \(\frac{d f}{\partial u}=f^{\prime} \text { and } \frac{d g}{\partial u}=g^{\prime} \text {. Now } \frac{\partial \mathrm{M}_1}{\partial y}=\frac{1}{x}\left[\frac{(f-g) f^{\prime} \frac{\partial u}{\partial y}-f\left(f^{\prime} \frac{\partial u}{\partial y}-g^{\prime} \frac{\partial u}{\partial y}\right)}{(f-g)^2}\right]\)
= \(\frac{1}{x} \frac{\partial u}{\partial y}\left[\frac{f f^{\prime}-g f^{\prime}-f f^{\prime}+f g^{\prime}}{(f-g)^2}\right]=\frac{1}{x} \cdot x \cdot \frac{f g^{\prime}-g f^{\prime}}{(f-g)^2}=\frac{f g^{\prime}-g f^{\prime}}{(f-g)^2}\) [from(2)] …………………..(3)
⇒ \(\frac{\partial \mathrm{N}_1}{\partial x}=\frac{1}{y}\left[\frac{(f-g) g^{\prime} \frac{\partial y}{\partial x}-g\left(f^{\prime} \frac{\partial u}{\partial x}-g^{\prime} \frac{\partial u}{\partial x}\right)}{(f-g)^2}\right]=\frac{1}{y} \frac{\partial u}{\partial x}\left[\frac{f g^{\prime}-g g^{\prime}-g f^{\prime}+g g^{\prime}}{(f-g)^2}\right]=\frac{f g^{\prime}-g f^{\prime}}{(f-g)^2}\) …………………….(4)
From (3) and (4): \(\frac{\partial \mathrm{M}_1}{\partial y}=\frac{\partial \mathrm{N}_1}{\partial x} \Rightarrow \mathrm{M}_1 d x+\mathrm{N}_1 d y=0\) is an exact equation
⇒ \(\frac{\mathrm{M}}{\mathrm{M} x-\mathrm{Ny}} d x+\frac{\mathrm{N}}{\mathrm{M} x-\mathrm{Ny}} d y=0\) is an exact equation.
Hence \(\frac{1}{M x-N y}\) is an integrating factor of \(\mathrm{M} d x+\mathrm{N} d y=0\)
Note. If Mx-Ny = 0, then M/N = y/x.
Then the equation reduces to y dx + x dy = 0 and its solution is xy = c.
Differential Equations of First Order and First Degree Working Rule to solve Mdx +Ndy = 0
1. General equation is Mdx + Ndy = 0 …. (1). Observe \(\frac{\partial \mathrm{M}}{\partial y} \neq \frac{\partial \mathrm{N}}{\partial x}\) => (1) is not exact.
2. Observe (1) is of the form \(y f(x y) d x+x g(x y) d y=0\)
3. Find Mx-Ny and observe it ≠ 0. Then \(\frac{1}{M x-N y}\) is an I.F of (1)
4. Multiply (1)with I.F. to transform it into an exact equation of (1) Mdx+ Ndy = 0 …………….. (2)
5. Solve (2) to get the general solution of (1)
Differential Equations of First Order and First Degree Method 4 To Find An Integrating Factor Of Mdx+Ndy=0
Definition And Types Of First-Order Differential Equations
Theorem 4. If there exists a continuous single variable function f (x) such that \(\frac{1}{N}\left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right)\) = f(x) or K (real number), then \(e^{\int f(x) d x}\) or \(e^{\int k d x}\) is an integrating factor of Mdx+Ndy=0
Proof. Given: \(\mathrm{M} d x+\mathrm{N} d y=0 \text { and } \frac{\partial \mathrm{M}}{\partial y}-\frac{\partial \mathrm{N}}{d x}=\mathrm{N} f(x)\) ……………………..(1)
Let us take \(\mathrm{M}_1=\mathrm{M} \exp \left(\int f(x) d x\right), \quad \mathrm{N}_1=\mathrm{N} \exp \left(\int f(x) d x\right)\)
Now, \(\frac{\partial M_1}{\partial y}-\frac{\partial \mathrm{N}_1}{\partial x}=\frac{\partial}{\partial y}\left[M \exp \left(\int f(x) d x\right)\right]-\frac{\partial}{\partial x}\left[N \exp \left(\int f(x) d x\right)\right]\)
= \(\exp \left(\int f(x) d x\right) \frac{\partial \mathrm{M}}{\partial y}-\exp \left(\int f(x) d x\right) \frac{\partial \mathrm{N}}{\partial x}-\mathrm{N} \exp \left(\int f(x) d x\right) \cdot \frac{\partial}{\partial x}\left(\int f(x) d x\right)\)
= \(\exp \left(\int f(x) d x\right) \frac{\partial \mathrm{M}}{\partial y}-\exp \left(\int f(x) d x\right) \frac{\partial \mathrm{N}}{\partial x}-\mathrm{N} f(x) \exp \left(\int f(x) d x\right)\)
= \(\exp \left(\int f(x) d x\right)\left[\left(\frac{\partial \mathrm{M}}{\partial y}-\frac{\partial \mathrm{N}}{\partial x}\right)-\mathrm{N} f(x)\right]=\exp \left(\int f(x) d x\right)[\mathrm{N} f(x)-\mathrm{N} f(x)]\)
= \(0 \Rightarrow \frac{\partial \mathrm{M}_1}{\partial y}-\frac{\partial \mathrm{N}_1}{\partial x}\) [From (1) ]. \(\mathrm{M}_1 d x+\mathrm{N}_1 d y=0\) is an exact equation.
⇒ \(\exp \left(\int f(x) d x\right)(\mathrm{M} d x+\mathrm{N} d y)=0\) is an exact equation.
Hence exp \(\left(\int f(x) d x\right)\) is an integrating factor of Mdx + NSy = 0.
Note 1. \(e^{\log f(x)}=f(x) \text { and } e^{k \log x}=e^{\log x^k}=x^k\) is a function of x alone or a real number k, then exp \(\left(\int f(x) d x\right)\) is an integrating factor of M dx + N dy = 0.
2. \(e^{\log f(x)}=f(x) \text { and } e^{k \log x}=e^{\log x^k}=x^k\) where k is constant.
First Order Differential Equations Exact Type Explained
Differential Equations of First Order and First Degree Working Rule to solve Mdx + Ndy = 0
1. General equation is \(\mathrm{M} d x+\mathrm{N} d y=0\) …. (1). Observe \(\frac{\partial \mathrm{M}}{\partial y} \neq \frac{\partial \mathrm{N}}{\partial x}\) => (1) is not exact.
2. Find \(\frac{1}{\mathrm{~N}}\left(\frac{\partial \mathrm{M}}{\partial y}-\frac{\partial \mathrm{N}}{\partial x}\right)\) and observe it as a function of x alone = f(x) or a real constant k.
3. Then \(e^{\int f(x) d x} \text { or } e^{\int k d x}\) is an I.F of (1)
4. Multiply (1)with I.F. to transform it into an exact equation of (1), \(\mathrm{M}_1 d x+\mathrm{N}_1 d y=0\) … (2)
5. Solve (2) to get the general solution of (1)
Differential Equations of First Order and First Degree Method 5 To Find An Integrating Factor Of Mdx + Ndy = 0
Solved Problems On First-Order And First-Degree Differential Equations
Theorem 5. If there exists a continuous and differentiable single variable function of g(y) such that \(\frac{1}{M}\left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)\) = g(y) or a real number k then \(\exp \left(\int g(y) d y\right) \quad o r \exp e x p\left(\int d y\right)\) is an integrating factor of Mdx+Ndy-0.
Proof. It is similar to the proof of Theorem 4.
Note. If \(\frac{1}{\mathrm{M}}\left(\frac{\partial \mathrm{N}}{\partial x}-\frac{\partial \mathrm{M}}{\partial y}\right)\) is a function of y alone, then \(\exp \left(\int g(y) d y\right)\) is an integrating factor of Mdx + Ndy = 0.
Differential Equations of First Order and First Degree Working Rule to solve Mdx + Ndy = 0
1. General equation is Mdx + Ndy = 0 …. (1). Observe \(\frac{\partial \mathrm{M}}{\partial y} \neq \frac{\partial \mathrm{N}}{\partial x}\)=> (1) is not exact
2. Find \(\frac{1}{\mathrm{M}}\left(\frac{\partial \mathrm{N}}{\partial x}-\frac{\partial \mathrm{M}}{\partial y}\right)\) and observe it is a function of y alone = g(y).
3. Then \(e^{\int g(y) d y}\) is an I.F of (1)
4. Multiply (1) with l.F. to transform it into an exact equation of (1), \(\mathrm{M}_1 d x+\mathrm{N}_{\mathrm{l}} d y=0\) … (2)
5. Solve (2) to get the general solution of (1)
Differential Equations of First Order and First Degree Method 6 to find an integrating factor of M dx + N dy = 0.
Step-By-Step Guide To Solving Exact Differential Equations
Theorem 6: If the differential equation Mdx+Ndy = 0 is of the form \(\left(m x^a y^b+p x^c y^d\right) y d x+\left(n x^a y^b+q x^c y^d\right) x d y=0\) (a,b,c,d,m,n, p are constants), then \(x^h y^k\) (h,k are constants) is an integrating factor of Mdx + Ndy = 0.
Proof. Let \(x^h y^k\) be the integrating factor of Mdx + Ndy = 0.
∴ \(x^h y^k\left[x^a y^b(m y d x+n x d y)+x^c y^d(p y d x+q x d y)\right]=0\)
⇒ \(x^h y^k\left[\left(m x^a y^b+p x^c y^d\right) y d x+\left(n x^a y^b+q x^c y^d\right) x d y\right]=0\)
⇒ \(\left(m x^{a+h} y^{b+k+1}+p x^{c+h} \cdot y^{d+k+1}\right) d x+\left(n x^{a+h+1} \cdot y^{b+k}+q x^{c+h+1} \cdot y^{d+k}\right) d y=0\) …….(1)
Here \(\mathrm{M}_1=m x^{a+h} y^{b+k+1}+p x^{c+h} y^{d+k+1}, \mathrm{~N}_1=n x^{a+h+1} y^{b+k}+q x^{c+h+1} y^{d+k}\)
⇒ \(\frac{\partial \mathrm{M}_1}{\partial y}=m(b+k+1) x^{a+h} y^{b+k}+p(d+k+1) x^{c+h} y^{d+k}\)
⇒ \(\frac{\partial \mathrm{N}_1}{\partial x}=n(a+h+1) x^{a+h} y^{b+k}+q(c+h+1) x^{c+h} y^{d+k}\)
(1) is exact if \(\frac{\partial \mathrm{M}_1}{\partial v}=\frac{\partial \mathrm{N}_1}{\partial x} \Rightarrow m(b+k+1)=n(a+h+1), p(d+k+1)=q(c+h+1)\)
⇒ \(n h-m k=m(b+1)-n(a+1)\) …………………….(2)
q h-p k=p(d+1)-q(c+1) ………………….(3)
Solving (2)and (3), we get the values of h and k.
Differential Equations of First Order and First Degree Working Rule to solve Mdx +Ndy = 0
1. General equation is Mdx + Ndy = 0 ….(1). Observe that \(\frac{\partial \mathrm{M}}{\partial \nu} \neq \frac{\partial \mathrm{N}}{\partial x} \Rightarrow\) (1) is not exact
2. Observe (1) is of the form \(x^a y^b(m y d x+n x d y)+x^c y^d(p y d x+q x d y)=0\) ………………..(2)
3. . Multiply the equation (2) with \(x^h y^k\)
4. After multiplication with \(x^h y^k\) the given equation (1) is exact, where \(\mathrm{M}_1\) = coefficient of dx, \(\mathrm{N}_1\) = coeff. of dy.
5. Find the values of h and k after using the condition \(\frac{\partial \mathrm{M}_1}{\partial y}=\frac{\partial \mathrm{N}_1}{\partial x}\)
6. Substitute the values of h and k in the transformed equation and then solve the
transformed equation.
Differential Equations of First Order and First Degree Linear Differential Equations Of First Order
Definition: An equation of the form \(\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}\) where P and Q are Constants or functions of x defined over an interval I alone is called a linear differential equation of first order in y.
If Q = 0 for all x in 1, then the corresponding equation \(\frac{d y}{d x}+\mathrm{P} y=0\) is called a homogeneous linear equation of first order.
If Q ≠ 0 for all x in I, then \(\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}\) is Called a non-homogeneous linear equation of first order.
Exact and linear First-Order Differential Equations With Examples
Theorem 7: If P and Q are differentiable functions of x Over an interval I then \(y \exp \left(\int P d x\right)=\int\left[Q \exp \left(\int \mathrm{P} d x\right)\right] d x+c\) is the general solution of the equation id \(\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}\)
Proof: The given equation is \(\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}\) …..(1) where P and Q are functions of x.
Rewriting (1) in the form Mdx +Ndy = 0, we get (Py-Q)dx+dy = 0 where M = Py-Q, N = 1 …………………………(2)
⇒ \(\frac{\partial \mathrm{M}}{\partial y}=\mathrm{P} \text { and } \frac{\partial \mathrm{N}}{\partial x}=0 \Rightarrow \frac{\partial \mathrm{M}}{\partial y} \neq \frac{\partial \mathrm{N}}{\partial x} \Rightarrow\) (2) is not an exact equation.
But \(\frac{1}{\mathrm{~N}}\left(\frac{\partial \mathrm{M}}{\partial y}-\frac{\partial \mathrm{N}}{\partial x}\right)=1(\mathrm{P}-0)=\mathrm{P}\) = f(x) = function of x alone
∴ By the method of 4 : \(e^{\int \mathrm{P} d x}\) is an integrating factor of (2) and hence in an I.F. of (1) Multiplying (1) by \(e^{\int P d x}\) , We get :
⇒ \(\exp \left(\int \mathrm{P} d x\right) \frac{d y}{d x}+\mathrm{P} \exp \left(\int \mathrm{P} d x\right) \cdot y=\mathrm{Q} \exp \left(\int \mathrm{P} d x\right) \Rightarrow \frac{d}{d x}\left[y \exp \left(\int \mathrm{P} d x\right)\right]=\mathrm{Q} \exp \left(\int \mathrm{P} d x\right)\)
Integrating: \(\int \frac{d}{d x}\left[y \exp \left(\int \mathrm{P} d x\right)\right] d x=\int\left[\mathrm{Q} \exp \left(\int \mathrm{P} d x\right)\right] d x+c\)
⇒ \(y \exp \left(\int \mathrm{P} d x\right)=\int\left[\mathrm{Q} \exp \left(\int \mathrm{P} d x\right)\right] d x+c\) is the general solution of \(\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}\)
Differential Equations of First Order and First Degree Working Rule to solve linear equation \(\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}\) where \(\mathrm{P}=f(x) \text { and } \mathrm{Q}=g(x)\)
1. First reduce the given equation to the standard form and then identify P and Q.
2. Find \(\int \mathrm{P} d x\) and then \(\text { I.F. }=e^{\int \mathrm{P} d x}\)
3. Then obtain a general solution by using: \(y(\text { I.F. })=\mathrm{Q}(\text { I.F) } d x+c\)
Note: \(e^{m \log \mathrm{A}}=\mathrm{A}^m \text { and } e^{-m \log \mathrm{A}}=1 / \mathrm{A}^m\) will be often used in simplifying I.F.
Differential Equations of First Order and First Degree Working rule for solving linear equation \(\frac{d x}{d y}+P_1 x=Q_1\)
1. First reduce the given equation to the standard form and then identify P₁ and Q₁.
2. Find \(\int \mathrm{P}_1 d y\) and then \(\text { I.F. }=\exp \left[\int \mathrm{P}_1 d y\right]\)
3. Then obtain G. S. by using \(x(\text { I.F. })=\int \mathrm{Q}_1 \text { (I.F) } d y+c\)
Equations Reducible To Linear Form
Differential Equations of First Order and First Degree Bernoulli’s Equation
An equation of the form \(\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q} y^n\) where P and Q are real numbers or functions of x alone and n is a real number such that n ≠ 0 and n ≠ 1, is called a Bernoulli’s differential equation. We can solve Bernoulli’s equation by reducing it to a linear differential equation in y as follows:
Solution. Given equation is \(\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q} y^n\) ………(1)
Multiplying (1) by \(y^{-n}\) we get: \(y^{-n} \frac{d y}{d x}+\mathrm{P} y^{-n+1}=\mathrm{Q}\)
Let \(y^{1-n}=u \Rightarrow(1-n) y^{-n} \frac{d y}{d x}=\frac{d u}{d x} \Rightarrow y^{-n} \frac{d y}{d x}=\frac{1}{1-n} \frac{d u}{d x}\) …………(3)
⇒ \(\frac{1}{1-n} \frac{d u}{d x}+\mathrm{P} u=\mathrm{Q} \Rightarrow \frac{d u}{d x}+(1-n) \mathrm{P} u=(1-n) \mathrm{Q}\) …………(4)
(4) is a linear equation of first order in u and x.
∴ \(\text { I.F. }=\exp \left[\int(1-n) \mathrm{P} d x\right]\). Hence the general solution of (4) is
u \(\exp \left[\int(1-n) \mathrm{P} d x\right]=(1-n) \int \mathrm{Q} \cdot \exp \left[\int(1-n) \mathrm{P} d x\right] d x+c\) ……….(5)
Substitution of \(u=y^{1-n}\) in (5), we get the general solution of (1).
Note : \(\frac{d x}{d y}+\mathrm{P}_1 x=\mathrm{Q}_1 x^n\) is also in the Bernoulli’s form, where \(\mathrm{P}_1 \text { and } \mathrm{Q}_1\) are functions of y only.
The method of solution is similar to that of the form (1) given above.
Differential Equations of First Order and First Degree Change Of Variables :
If the given differential equation does not directly come under any of the forms discussed so far to one of these forms, we can reduce the given differential equation by suitable substitution. This procedure of reducing the given differential equation by substitution is called the change of dependent or independent variable.
Equations Reducible To First Order And First Degree By \(p=\frac{d y}{d x}\) Substitution.
Consider the differential equation \(f\left(\frac{d^2 y}{d x^2}, \frac{d y}{d x}, x\right)=0\) not containing y directly.
By putting \(\frac{d y}{d x}=p\) the equation can be transformed as \(F\left(\frac{d p}{d x}, p, x\right)=0\) which is of first order and first degree.
Applications Of First-Order Differential Equations In Real Life
Example. 1 Solve \(x \frac{d^2 y}{d x^2}+\frac{d y}{d x}+1=0\)
Solution.
Given
\(x \frac{d^2 y}{d x^2}+\frac{d y}{d x}+1=0\)Put \(\frac{d y}{d x}=p \text { so that } \frac{d^2 y}{d x^2}=\frac{d p}{d x}\)
∴ The given equation takes the form \(x \frac{d p}{d x}+p+1=0 \Rightarrow x \frac{d p}{d x}=-(p+1)\)
⇒ \(\frac{1}{p+1} d p=-\frac{1}{x} d x\)
∴ \(\int \frac{1}{p+1} d p=-\frac{1}{x} \cdot d x\)
∴ \(\log (p+1)=\log \frac{c}{x}\)
∴ \(p+1=\frac{c}{x} \Rightarrow \frac{d y}{d x}=\frac{c}{x}-1 \Rightarrow d y=\left(\frac{c}{x}-1\right) d x\)
Integrating the general solution is \(y=c \log x-x+c\)
Homogeneous And Non-Homogeneous First-Order Differential Equations
Example. 2: Solve \(\frac{d^2 y}{d x^2}+\frac{d y}{d x} \tan x=\sec x+\cos x\)
Solution.
Given
\(\frac{d^2 y}{d x^2}+\frac{d y}{d x} \tan x=\sec x+\cos x\)Put \(\frac{d y}{d x}=p\) so that \(\frac{d^2 y}{d x^2}=\frac{d p}{d x}\)
∴ The given equation takes the form \(\frac{d p}{d x}+p \tan x=\sec x+\cos x\) which is linear equation.
∴ I.F. = \(e^{f \tan x d x}=e^{\log \sec x}=\sec x\)
G. S. is P (I.F.) = \(\mathrm{Q} \int(\mathrm{I} . \mathrm{F}) d x+c\)
∴ \(p \sec x=\int \sec x(\sec x+\cos x) d x+c=\tan x+x+c\)
∴ \(\frac{d y}{d x}=\sin x+x \cos x+c \cos x\)
∴ The general solution is y = \(\int(\sin x+c \cos x+x \cos x) d x+c^{\prime}\)
⇒ \(y=-\cos x+c \sin x+(x \sin x+\cos x)+c^{\prime} \Rightarrow y=x \sin x+c \sin x+c^{\prime}\)