Differential Equations of First Order and First Degree Solved Problems
Example. 1. Solve \(\left(x y^2-x^2\right) d x+\left(3 x^2 y^2+x^2 y-2 x^3+y^2\right) d y=0\)
Solution.
Given equation is \(\left(x y^2-x^2\right) d x+\left(3 x^2 y^2+x^2 y-2 x^3+y^2\right) d y=0\) ……………..(1)
where \(\mathrm{M}=x y^2-x^2, \mathrm{~N}=3 x^2 y^2+x^2 y-2 x^3+y^2\)
⇒ \(\frac{\partial \mathrm{M}}{\partial y}=2 x y, \frac{\partial \mathrm{N}}{\partial x}\)
= \(6 x y^2+2 x y-6 x^2 \Rightarrow \frac{\partial \mathrm{M}}{\partial y} \neq \frac{\partial \mathrm{N}}{\partial x}\) => (1) is not an exact equation.
But \(\frac{1}{M}\left(\frac{\partial \mathrm{N}}{\partial x}-\frac{\partial \mathrm{M}}{\partial y}\right)\)
= \(\frac{1}{x y^2-x^2}\left(6 x y^2+2 x y-6 x^2-2 x y\right)=\frac{1}{x y^2-x^2} \cdot 6\left(x y^2-x^2\right)=6K\)
∴ I.F = \(e^{\int 6 d y}=e^{6 y}\) Multiplying (1) by \(e^{6 y} \Rightarrow\left(x y^2-x^2\right) e^{6 y} d x+\left(3 x^2 y^2+x^2 y-2 x^3+y^2\right) e^{6 y} d y=0\) …………………(2)
(2) is an exact equation where \(\mathrm{M}_1=\left(x y^2-x^2\right) e^{6 y} \text { and } \mathrm{N}_1=\left(3 x^2 y^2+x^2 y-2 x^3+y^2\right) e^{6 y}\) since \(\frac{\partial \mathrm{M}_1}{\partial y}=\frac{\partial \mathrm{N}_1}{\partial x}\) (verify)
(1) Integrating \(\mathrm{M}_1\) w.r.t. x, treating y as constant.
⇒ \(\int^x \mathrm{M}_1 d x=\int^x\left(x y^2-x^2\right) e^{6 y} d x=e^{6 y}\left(\frac{x^2}{2} y^2-\frac{x^3}{3}\right)\) …………………(3)
(2) \(\int\) (terms of \(\mathrm{N}_1\) not containing x) dy = \(\int y^2 e^{6 y} d y\)
= \(\frac{y^2 e^{6 y}}{6}-\int \frac{e^{6 y}}{6} \cdot 2 y d y=y^2 \frac{e^{6 y}}{6}-\frac{1}{3}\left[y \frac{e^{6 y}}{6}-\int \frac{e^{6 y}}{6} \cdot d y\right]\)
= \(\frac{y^2 e^{6 y}}{6}-\frac{1}{18} y e^{6 y}+\frac{1}{18} \frac{e^{6 y}}{6}\) ……………..(4)
∴ The general solution of (2) is (3) + (4) = C
⇒ \(\frac{x^2 y^2 e^{6 y}}{2}-\frac{x^3 e^{6 y}}{3}+\frac{y^2 e^{6 y}}{6}-\frac{y e^{6 y}}{18}+\frac{e^{6 y}}{108} \neq c \Rightarrow e^{6 y}\left(\frac{x^2 y^2}{2}-\frac{x^3}{3}+\frac{y^2}{6}-\frac{y}{18}+\frac{1}{108}\right)=c\)
Differential Equations Of First Order And First Degree Exercise 2.5
Example 2. Solve \(\left(x y^3+y\right) d x+2\left(x^2 y^2+x+y^4\right) d y=0\)
Solution.
Given equation is of the form Mdx + Ndy = 0
where \(\mathrm{M}=x y^3+y \text { and } \mathrm{N}=2\left(x^2 y^2+x+y^4\right)\)
Now, \(\frac{\partial \mathrm{M}}{\partial y}=3 x y^2+1, \frac{\partial \mathrm{N}}{\partial x}=4 x y^2+2 \Rightarrow \frac{\partial \mathrm{M}}{\partial y} \neq \frac{\partial \mathrm{N}}{\partial x} \Rightarrow\)
Given equation is not an exact equation
But \(\frac{1}{M}\left(\frac{\partial \mathrm{N}}{\partial x}-\frac{\partial \mathrm{M}}{\partial y}\right)=\frac{1}{x y^3+y}\left(4 x y^2+2-3 x y^2-1\right)=\frac{x y^2+1}{y\left(x y^2+1\right)}=\frac{1}{y}=g(y)\)
∴ \(\text { I.F. }=\exp \left(\int \frac{1}{y} d y\right)=\exp (\log y)=e^{\log y}=y\)
Multiplying the given equation with \(y: \Rightarrow\left(x y^4+y^2\right) d x+2\left(x^2 y^3+x y+y^5\right) d y=0\) ……(1)
Then (1) is an exact equation where \(\mathrm{M}_1=x y^4+y^2\)
and \(\mathrm{N}_1=2\left(x^2 y^3+x y+y^5\right) \Rightarrow \frac{\partial \mathrm{M}_1}{\partial y}=\frac{\partial \mathrm{N}_1}{\partial x}\) (verify)
(1) Integrating \(\mathrm{M}_1\) w.r.t. x, treating y as constant
⇒ \(\int^x \mathrm{M}_1 d x=\int^x\left(x y^4+y^2\right) d x=\frac{x^2 y^4}{2}+y^2 x\) ………………………..(2)
(2) \(\int\) (terms of \(\mathrm{N}_1\) not containing x) dy = \(\int 2 y^5 d y=\frac{y^6}{3}\) …………….(3)
∴ The general solution of (1) is (2) + (3) = C
⇒ \(\frac{x^2 y^4}{2}+y^2 x+\frac{y^6}{3}=c \Rightarrow 3 x^2 y^4+6 x y^2+2 y^6=6 c\)
Homogeneous Equations Solved Problems Exercise 2.5
Example. 3. Solve \(\left(y^4+2 y\right) d x+\left(x y^3+2 y^4-4 x\right) d y=0\)
Solution.
Given equation is \(\left(y^4+2 y\right) d x+\left(x y^3+2 y^4-4 x\right) d y=0\) ………………………..(1)
Where \(\mathrm{M}=y^4+2 y, \mathrm{~N}=x y^3+2 y^4-4 x \Rightarrow \frac{\partial \mathrm{M}}{\partial v}=4 y^3+2, \frac{\partial \mathrm{N}}{\partial x}=y^3-4\)
Since, \(\frac{\partial \mathrm{M}}{\partial y} \neq \frac{\partial \mathrm{N}}{\partial x}\) (1) is not an exact equation
But \(\frac{1}{\mathrm{M}}\left(\frac{\partial \mathrm{N}}{\partial x}-\frac{\partial \mathrm{M}}{\partial y}\right)=\frac{1}{y\left(y^3+2\right)}\left(y^3-4-4 y^3-2\right)=\frac{-3\left(y^3+2\right)}{y\left(y^3+2\right)}=\frac{-3}{y}\) Tan-1
∴ I.F = \(\exp \left(\int \frac{-3}{y} d y\right)=\exp (-3 \log y)=\exp \left(\log y^{-3}\right)=\frac{1}{y^3}\)
Multiplying (1) with \(\frac{1}{y^3} \Rightarrow\left(y+\frac{2}{y^2}\right) d x+\left(x+2 y-\frac{4 x}{y^3}\right) d y=0\) ……………………..(2)
(2) is an exact equation where \(\mathrm{M}_1=y+\frac{2}{y^2}, \mathrm{~N}_1=x+2 y-\frac{4 x}{y^3}\)
∴ The general solution of (2) is \(\int^x\left(y+\frac{2}{y^2}\right) d x+\int 2 y d y=\mathrm{C} \Rightarrow\left(y+\frac{2}{y^2}\right) x+y^2=\mathrm{c}\)
Differential Equations Of First Order And First Degree Exercise 2.5
Example.4. Solve \(\left(2 x^2 y-3 y^2\right) d x+\left(2 x^3-12 x y+\log y\right) d y=0\)
Solution.
Given equation is of the form M dx + N dy = 0 …………………(1)
where \(\mathrm{M}=2 x^2 y-3 y^2, \mathrm{~N}=2 x^3-12 x y+\log y\)
⇒ \(\frac{\partial \mathrm{M}}{\partial y}=2 x^2-6 y, \frac{\partial \mathrm{N}}{\partial x}=6 x^2-12 y \Rightarrow \frac{\partial \mathrm{M}}{\partial y} \neq \frac{\partial \mathrm{N}}{\partial x}\)
=> The given equation is not an exact equation.
But \(\frac{1}{\mathrm{M}}\left(\frac{\partial \mathrm{N}}{\partial x}-\frac{\partial \mathrm{M}}{\partial y}\right)=\frac{1}{2 x^2 y-3 y^2}\left(6 x^2-12 y-2 x^2+6 y\right)=\frac{2\left(2 x^2-3 y\right)}{y\left(2 x^2-3 y\right)}=\frac{2}{y}\)
∴ I.F \(=e^{\int(2 / y) d y}=e^{2 \log y}=e^{\log y^2}=y^2\)
Multiplying (1) by \(y^2:\left(2 x^2 y^3-3 y^4\right) d x+\left(2 x^3 y^2-12 x y^3+y^2 \log y\right) d y=0\) ……………..(2)
(2) is an exact equation where \(\mathrm{M}_1=2 x^2 y^3-3 y^4, \quad \mathrm{~N}_1=2 x^3 y^2-12 x y^3+y^2 \log y\)
⇒ \(\frac{\partial \mathrm{M}_1}{\partial y}=6 x^2 y^2-12 y^3=6 x^2 y^2-12 y^3=\frac{\partial \mathrm{N}_1}{\partial x}\) ……………………(3)
(1) Integrating M1 w.r.t.x., treating y as constant.
⇒ \(\int^x M_1 d x=\int^x\left(2 x^2 y^3-3 y^4\right) d x=(2 / 3) x^3 y^3-3 y^4 x\)
(2) \(\int\) (terms of \(\mathrm{N}_1\) not containing x)dy = \(\int y^2 \log y d y\)
= \((1 / 3) y^3 \log y-\int(1 / 3) y^3 \cdot(1 / y) d y=(1 / 3) y^3 \log y-(1 / 3) y^2 d y\)
= \(\frac{1}{3} y^3 \log y-\frac{1}{27} y^3\) …………………….(4)
The general solution (2) is (3) + (4) = c \(\frac{2}{3} x^3 y^3-3 y^4 x+\frac{y^3}{3} \log y-\frac{1}{9} y^3=\frac{c}{9} \Rightarrow 6 x^3 y^3-27 x y^4+3 y^3 \log y=c\)