Differential Equations Introduction Solved Problems
Example. 1: Solve : \(x^2 y d x-\left(x^3+y^3\right) d y=0\)
Solution:
Given: \(x^2 y d x=\left(x^3+y^3\right) d y \Rightarrow \frac{d y}{d x}=\frac{x^2 y}{x^3+y^3}\) …………………..(1)
Since \(f(k x, k y)=f(x, y)\), (1) is homogeneous equation.
Put \(y / x=v \text { in (1) } \Rightarrow y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\) ……………………(2)
(1) and (2) => \(v+x \frac{d v}{d x}=\frac{v}{1+v^3} \Rightarrow x \frac{d v}{d x}=\frac{v}{1+\dot{v}^3}-v=\frac{-v^4}{1+v^3}\)
Separating the variables: \(\frac{1+v^3}{v^4} d v=-\frac{d x}{x}\)
Integrating: \(\int \frac{1+v^3}{v^4} d v=-\int \frac{d x}{x}+c \Rightarrow \int\left(v^{-4}+\frac{1}{v}\right) d v=-\int \frac{d x}{x}+c\)
⇒ \(\frac{v^{-3}}{-3}+\log v=-\log x+c \Rightarrow \log v-\frac{1}{3 v^3}+\log x=c\) …………………….(3)
Putting v = y/x in (3), the general solution is \(\log (y / x)-\left(x^3 / 3 y^3\right)+\log x=c\)
Solved Problems On First Order And First-Degree Homogeneous Differential Equations
Example. 2: Solve : \((x-y \log y+y \log x) d x+x(\log y-\log x) d y=0\)
Solution:
The given equation is
x\(\log \left(\frac{y}{x}\right) d y+\left[x-y \log \left(\frac{y}{x}\right)\right] d x=0 \Rightarrow \frac{d y}{d x}=\frac{y \log (y / x)-x}{x \log (y / x)}=\frac{(y / x) \log (y / x)-1}{\log (y / x)}\) ………………(1)
Clearly \(\frac{d y}{d x}=f\left(\frac{y}{x}\right)\) => Given equation.is homogeneous.
Put y/x = v in (1) \(y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\) …………………..(2)
(1) and (2) \(\Rightarrow v+x \frac{d v}{d x}=\frac{v \log v-1}{\log v} \Rightarrow x \frac{d v}{d x}=-\frac{1}{\log v}\)
Separating the variables: \((\log v) d v=-\frac{d x}{x}\)
⇒ \(\int \log v \cdot d v=\int-\frac{d x}{x}+c \Rightarrow v \log v-\int v \cdot \frac{1}{v} d v=-\log x+c\)
⇒ \(v \log v-v=-\log x+c\) …………………….(3)
Put v = y/x in (3)
∴ The general solution of (1) is (y/x) log(y/x)-(y/x) = -log x + c
=> y (log y – log x) – y = -x log x + cx => y log y + (x – y) log x = y + cx
Example. 3. Solve : \(x \frac{d y}{d x}=y+x e^{y / x}\)
Solution.
Given equation is \(x \frac{d y}{d x}=y+x e^{y / x} \Rightarrow \frac{d y}{d x}=\frac{y}{x}+e^{y / x}\) …………………(1)
∵ \(f(k x, k y)=f(x, y)\), (1) is a homogeneous equation.
Put \(y=v x \text { in }(1) \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\) ……………………..(2)
Then (1) and (2) \(\Rightarrow v+x \frac{d v}{d x}=v+e^v \Rightarrow x \frac{d v}{d x}=e^v\)
Separating the variables: \(\frac{d v}{e^v}=\frac{d x}{x} \Rightarrow e^{-v} d v=\frac{d x}{x}\)
Integrating : \(\int e^{-v} d v=\int \frac{d x}{x}+c \Rightarrow-e^{-v}=\log |x|+c\) …………………..(3)
Putting v = y / x in (3), the general solution of (1) is \(e^{-(y / x)}+\log |x|+c=0\)
Step-By-Step Guide To Homogeneous Differential Equations First Order
Example. 4. Solve : \(x d y-y d x=\left(\sqrt{x^2+y^2}\right) d x\)
Solution.
Given equation is \(x d y-y d x=\left(\sqrt{x^2+y^2}\right) d x\)
⇒ \(x \frac{d y}{d x}=y+\sqrt{x^2+y^2} \Rightarrow \frac{d y}{d x}=\frac{y+\sqrt{x^2+y^2}}{x}\) …………………(1)
∵ \(f(k x, k y)=f(x, y)\), (1) is a homogeneous equation.
Put y/x = v in (1) y = vx in (1) \(\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\) …………………….(2)
(1) and (2) ⇒ \(v+x \frac{d v}{d x}=v+\sqrt{1+v^2} \Rightarrow \frac{x d v}{d x}=\sqrt{1+v^2}\)
Separating the variables: \(\frac{d v}{\sqrt{1+v^2}}=\frac{d x}{x}\)
Integrating: \(\int \frac{d v}{\sqrt{1+v^2}}=\int \frac{d x}{x}+\log c \Rightarrow \log \left|v+\sqrt{1+v^2}\right|=\log x+\log c\)
⇒ \(\log \left|v+\sqrt{1+v^2}\right|=\log c x \Rightarrow v+\sqrt{1+v^2}=c x\)
Putting v = y/x in (3), the general solution of (1) is \(y+\sqrt{x^2+y^2}=c x^2\)
Example. 5: Solve : [x – y arctan(y / x)] Jx + [x arctan(y / x)] dy = 0
Solution.
Given equation
[x – y arctan(y / x)] Jx + [x arctan(y / x)] dy = 0
The given equation is rearranged as follows: \(\frac{d y}{d x}=\frac{y \text{Tan}^{-1}(y / x)-x}{x \text{Tan}^{-1}(y / x)}\)
= \(\frac{(y / x) \text{Tan}^{-1}(y / x)-1}{\text{Tan}^{-1}(y / x)}\)…….(1)
because f(k x, k y)=f(x, y),(1) is homogeneous equation.
Put y=v x in (1) ⇒ \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)……(2)
(1) and (2) ⇒ \(v+x \frac{d v}{d x}=\frac{v \text{Tan}^{-1} v-1}{\text{Tan}^{-1} \cdot v} \Rightarrow x \frac{d v}{d x}\)
= \(\frac{v \text{Tan}^{-1} v-1}{\text{Tan}^{-1} v}-v \Rightarrow x \frac{d v}{d x}=\frac{-1}{\text{Tan}^{-1} v}\)
Separating the variables: \(\text{Tan}^{-1} v d v=-\frac{d x}{x}\).
Integrating: \(\int \text{Tan}^{-1} v d v=-\int \frac{d x}{x}+c\).
⇒ \(v \text{Tan}^{-1} v-\int \frac{v}{1+v^2} d v=-\log |x|+c\)
⇒ \(v \text{Tan}^{-1} v-\frac{1}{2} \log \left(1+v^2\right)=-\log |x|+c \ldots\)……(3)
Putting \(v=y / x\) in (3), the general solution of (1) is \(\frac{y}{x} \text{Tan}^{-1}\left(\frac{y}{x}\right)-\frac{1}{2} \log \left(1+\frac{y^2}{x^2}\right)=-\log |x|+c\)
Working Rules Of First Order Homogeneous Differential Equations With Examples
Example. 6. Solve \((y d x+x d y) x \cos (y / x)=(x d y-y d x) y \sin (y / x)\)
Solution.
Given : \(\left[x y \cos (y / x)+y^2 \sin (y / x)\right] d x=\left[x y \sin (y / x)-x^2 \cos (y / x)\right] d y\)
⇒ \(\frac{d y}{d x}=\frac{x y \cos (y / x)+y^2 \sin (y / x)}{x y \sin (y / x)-x^2 \cos (y / x)}\) ………………..(1)
f(k x, k y)=f(x, y), (1) is a homogeneous equation.
Put \(y / x=v \Rightarrow y=v x \text { in }(1) \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\) ……………………..(2)
(1) and (2) \(\Rightarrow v+x \frac{d v}{d x}=\frac{v \cos v+v^2 \sin v}{v \sin v-\cos v} \Rightarrow x \frac{d v}{d x}=\frac{v \cos v+v^2 \sin v}{v \sin v-\cos v}-v=\frac{2 v \cdot \cos v}{v \sin v-\cos v}\)
Separating the variables : \(2 \frac{d x}{x}=\frac{v \sin v-\cos v}{v \cos v} d v\)
Integrating : \(2 \int \frac{d x}{x}=\int \frac{v \sin v-\cos v}{v \cos v} d v+\log c\)
⇒ \(2 \log |x|=\int \tan v d v-\int \frac{1}{v} d v+\log c \Rightarrow 2 \log |x|=-\log |\cos v|-\log |v|+\log c\)
⇒ \(\log x^2=\log \left|\frac{c}{v \cos v}\right| \Rightarrow x^2 v \cos v=c \Rightarrow x^2 \cdot \frac{y}{x} \cos \frac{y}{x}=c\)
∴ The general solution of (1) is xy cos(y/x) = c
Example. 7 : Solve: \(\left(1+e^{x / y}\right) d x+e^{x / y}[1-(x / y)] d y=0\)
Solution.
Given equation is \(\left(1+e^{x / y}\right) d x+e^{x / y}[1-(x / y)] d y=0\)
Here we shall not be able to write the given equation in the form \(\frac{d y}{d x}=f\left(\frac{y}{x}\right)\).
But we can express it in the form \(\frac{d x}{d y}=f\left(\frac{x}{y}\right)\)
∴ Given equation is \(\frac{d x}{d y}=\frac{e^{x / y}[(x / y)-1]}{1+e^{x / y}}=f\left(\frac{x}{y}\right)\) ……………………(1)
f(kx, ky) = f(x, y), (1) is a homogeneous equation.
Put x/y = v where v is a function of y => x = vy => \(\frac{d x}{d y}=v+y \frac{d v}{d y}\) …………………..(2)
(1) and (2) => \(v+y \frac{d v}{d y}=\frac{e^v(v-1)}{1+e^v} \Rightarrow y \frac{d v}{d y}=\frac{e^v(v-1)}{1+e^v}-v=-\frac{e^v+v}{1+e^v}\)
Separating the variables: \(\frac{d y}{y}=-\frac{1+e^v}{v+e^v} d v\)
Integrating: \(\int \frac{d y}{y}=-\int \frac{1+e^v}{v+e^v} d v+\log c \Rightarrow \log |y|=-\log \left|v+e^v\right|+\log c\)
⇒ \(\log y+\log \left(v+e^v\right)=\log c \Rightarrow \log y\left(v+e^v\right)=\log c \Rightarrow y\left(v+e^v\right)=c\) …………………….(3)
Putting v = x/y in (3), the general solution of (1) is \(x+y e^{x / y}=c\)
Introduction To First Degree Homogeneous Differential Equations Tutorial
Example. 8: Solve \(x^2 \frac{d y}{d x}=\frac{y(x+y)}{2}\)
Solution:
Given equation is \(\frac{d y}{d x}=\frac{x y+y^2}{2 x^2}\) ……………………..(1)
∴ f(kx>ty) = f(x,y), (1) is a homogeneous equation.
Put y = vx in(1) \(\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\) ……………………(2)
(1),(2) ⇒ \(v+x \frac{d v}{d x}=\frac{v x^2+v^2 x^2}{2 x^2}=\frac{v^2+v}{2}\)
⇒ \(x \frac{d v}{d x}=\frac{v^2+v}{2}-v=\frac{v^2+v-2 v}{2}=\frac{v^2-v}{2}\)
Separating variables: \(2 \frac{d v}{v(v-1)}=\frac{d x}{x}\)
⇒ \(2 \int \frac{d v}{v(v-1)}=\int \frac{d x}{x}+\log c\)
2\( \int\left(\frac{1}{v-1}-\frac{1}{v}\right) d v=\log x+\log c \Rightarrow 2[\log (v-1)-\log v]=\log c x\)
⇒ \(\log \left(\frac{v-1}{v}\right)^2=\log (c x) \Rightarrow\left(\frac{v-1}{v}\right)^2=c x\)
Putting \(v=\frac{y}{x}\), the general solution of (1) is \(\left[\frac{(y / x)-1}{y / x}\right]^2=c x \Rightarrow(y-x)^2=c x y^2\)
Homogeneous Differential Equations Working Rules Step By Step
Example. 9: Find the equation of the curve, which passes through the point (1, π/4) whose differential equation is \(\left[x \cos ^2(y / x)-y\right] d x+x d y=0(x>0, y>0)\)
Solution.
Given \(\left[x \cos ^2(y / x)-y\right] d x+x d y=0\)
⇒ \(\frac{d y}{d x}=\frac{y-x \cos ^2(y / x)}{x}=\frac{y}{x}-\cos ^2\left(\frac{y}{x}\right)\) ……………………..(1)
Clearly, \(\frac{d y}{d x}=\mathrm{F}\left(\frac{y}{x}\right)\) => Given equation is homogeneous.
Put v = y/x in (1) => \(y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\) …………………..(2)
(1) and (2) \(\Rightarrow v+x \frac{d v}{d x}=v-\cos ^2 v \Rightarrow x \frac{d v}{d x}=-\cos ^2 v\).
Separating the variables: \(\sec ^2 v d v=-\frac{d x}{x}\)
⇒ \(\int \sec ^2 v d v=-\int \frac{d x}{x}+c \Rightarrow \tan v=-\log |x|+c\) ………………..(3)
Putting v = y/x in (3), the G. S. is tan (y/x) = —log 1 x | +c …………………………..(4)
Putting x = 1 and y = π/4 in (4):
tan (π / 4) = – log 1 + c tan (π / 4) = – log 1 + c => c = 1
The equation of the curve is tan(y/x) + log | x | = 1