Differential Equations Introduction Solved Problems
Existence And Uniqueness Theorem Solved Problems
Example. 1: Solve : \(\frac{d y}{d x}+\sqrt{\frac{1-y^2}{1-x^2}}=0\)
Solution.
Given equation is \(\frac{d y}{d x}+\sqrt{\frac{1-y^2}{1-x^2}}=0\)…..(1)
Separating the variables: \(\frac{d y}{\sqrt{1-y^2}}+\frac{d x}{\sqrt{1-x^2}}=0\)
⇒ \(\int \frac{d y}{\sqrt{1-y^2}}+\int \frac{d x}{\sqrt{1-x^2}}=c\) (c being arbitrary constant)
∴ The general solution of (1) is: \(\text{Sin}^{-1} y+\text{Sin}^{-1} x=c\)
Examples Of Solved Problems On Existence And Uniqueness Theorem
Example. 2: Solve : \(y \frac{d y}{d x}=x e^{x^2+y^2}\)
Solution.
Given equation is \(y \frac{d y}{d x}=x e^{x^2} \cdot e^{y^2}\) …………………(1)
Separating the variables we have : \(x e^{x^2} d x=y e^{-y^2} d y \Rightarrow \int x e^{x^2} d x=\int y e^{-y^2} d y\)
Put \(x^2=u, y^2=t \Rightarrow 2 x d x=d u, 2 y d y=d t\)
∴ \(\frac{1}{2} \int e^u d u=\frac{1}{2} \int e^{-t} d t+\frac{c}{2} \Rightarrow e^u=-e^{-t}+c\)
∴ The general solution of (1) is \(e^{x^2}+e^{-y^2}=c\)
Application Of Existence And Uniqueness Theorem In Differential Equations
Example. 3: Solve : \(\left(\frac{d y}{d x}\right) \tan y=\sin (x+y)+\sin (x-y)\)
Solution.
Given \(\left(\frac{d y}{d x}\right) \tan y=2 \sin x \cos y\) ………………..(1)
Separating the variables : \(\frac{\tan y}{\cos y} d y=2 \sin x d x\)
=> \(\int\) sec y tan y dy = \(\int\) 2 sin x dx => \(\sec y=-2 \cos x+c\)
∴ The general solution of (1) is sec y + 2 cos x = c
Solved Examples Of Initial Value Problems Using The Existence And Uniqueness Theorem
Example. 4: Solve: \(\log \left(\frac{d y}{d x}\right)=a x+b y\)
Solution:
Given equation is \(\log \left(\frac{d y}{d x}\right)=a x+b y\) ………………………(1)
⇒ \(\frac{d y}{d x}=e^{a x+b y}=e^{a x} \cdot e^{b y} \Rightarrow e^{a x} d x=e^{-b y} d y\)
Separating the variables => \(\int e^{a x} d x=\int e^{-b y} d y+\mathrm{c} \Rightarrow \frac{e^{a x}}{a}=\frac{e^{-b y}}{-b}+\mathrm{c}\)
∴ The given solution of (1) is \(b e^{a x}+a e^{-b y}=c\)
Problems On Existence And Uniqueness Theorem With Solutions
Example. 5: Solve : \(\frac{d y}{d x}=\frac{x(2 \log x+1)}{\sin y+y \cos y}\)
Solution:
Given equation is \(\frac{d y}{d x}=\frac{x(2 \log x+1)}{\sin y+y \cos y}\) ……………………..(1)
Separating the variables : x(2 log x +1) dx = (sin y + y cos y) dy
Integrating: \(2 \int x \log x d x+\int x d x=\int \sin y d y+\int y \cos y d y\)
⇒ \(2\left(\frac{x^2}{2} \log x-\int \frac{x^2}{2} \cdot \frac{1}{x} d x\right)+\frac{x^2}{2}=-\cos y+y \sin y-\int \sin y d y\)
⇒ \(x^2 \log x-\frac{x^2}{2}+\frac{x^2}{2}=-\cos y+y \sin y+\cos y+c \Rightarrow x^2 \log x=y \sin y+c\)
The general solution of (1) is \(x^2 \log x=y \sin y+c\)
Lipschitz Condition In Existence And Uniqueness Theorem Problems
Example. 6: Solve \(y-x \frac{d y}{d x}=a\left(y^2+\frac{d y}{d x}\right)\)
Solution:
Given equation is \(y-x \frac{d y}{d x}=a\left(y^2+\frac{d y}{d x}\right)\) ………………….(1)
⇒ \((a+x) \frac{d y}{d x}=y-a y^2\)
Separating the variables : \(\frac{1}{y-a y^2} d y=\frac{1}{a+x} d x \Rightarrow \int \frac{1}{y(1-a y)} d y=\int \frac{1}{a+x} d x+c_1\)
⇒ \(\int\left(\frac{1}{y}+\frac{a}{1-a y}\right) d y=\int \frac{1}{a+x} d x+c_1 \Rightarrow \log |y|-\log |1-a y|=\log \mid a+x \mid+\log c\)
⇒ \(\log \left|\frac{y}{1-a y}\right|=\log |c(a+x)| \Rightarrow \frac{y}{1-a y}=c(a+x)\)
∴ The general solution of (1) is y = c (a + x) (1 – ay)
Existence And Uniqueness Theorem Numerical Examples
Example. 7: Solve : \(3 e^x \tan y d x+\left(1-e^x\right) \sec ^2 y d y=0\)
Solution.
Given equation is \(3 e^x \tan y d x+\left(1-e^x\right) \sec ^2 y d y=0\) …………………..(1)
Separating the variables: \(\frac{3 e^x}{1-e^x} d x+\frac{\sec ^2 y}{\tan y} d y=0\)
⇒ \(3 \int \frac{e^x}{1-e^x} d x+\int \frac{\sec ^2 y}{\tan y} d y=\log c \quad \Rightarrow-3 \log \left|\left(1-e^x\right)\right|+\log |\tan y|=\log c\)
⇒ \(\log \left|\left(1-e^x\right)^{-3}\right|+\log |\tan y|=\log c \Rightarrow \log \left|\frac{\tan y}{\left(1-e^x\right)^3}\right|=\log c \Rightarrow \frac{\tan y}{\left(1-e^x\right)^3}=c\)
∴ The general solution of (1) is \(\tan y=c\left(1-e^x\right)^3\)
Differential Equations Problems Explained Using Existence And Uniqueness Theorem
Example. 8: Solve : \(\frac{d y}{d x}=(4 x+y+1)^2\)
Solution:
Given equation is \(\frac{d y}{d x}=(4 x+y+1)^2\)……..(1)
Let \(4 x+y+1=u \Rightarrow 4+\frac{d y}{d x}=\frac{d u}{d x} \Rightarrow \frac{d y}{d x}=\frac{d u}{d x}-4\)…..(2)
From (1) and (2): \(\frac{d u}{d x}-4=u^2 \Rightarrow \frac{d u}{d x}=u^2+4\).
Separating the variables: \(\frac{d u}{u^2+4}=d x\)
⇒ \(\int \frac{d u}{u^2+2^2}=\int d x+\frac{c}{2} \Rightarrow \frac{1}{2} \text{Tan}^{-1}\left(\frac{u}{2}\right)\)
= \(x+\frac{c}{2} \Rightarrow \text{Tan}^{-1}\left(\frac{u}{2}\right)=2 x+c\)
u = \(2 \tan (2 x+c)\)
∴ The General solution is \(4 x+y+1=2 \tan (2 x+c)\)
(because \(u=4 \dot{x}+y+1)\)
Worked Examples On Existence And Uniqueness Theorem In Differential Equations
Example 9: Solve : \(\frac{d y}{d x}=\cos (x+y)+\sin (x+y)\)
Solution:
Given equation is \(\frac{d y}{d x}=\cos (x+y)+\sin (x+y)\) …………………..(1)
Let \(x+y=u \Rightarrow 1+\frac{d y}{d x}=\frac{d u}{d x} \Rightarrow \frac{d y}{d x}=\frac{d u}{d x}-1\) ……………………..(2)
(1)and (2) => \(\frac{d u}{d x}-1=\cos u+\sin u \Rightarrow \frac{d u}{d x}=(1+\cos u)+\sin u\)
⇒ \(\frac{d u}{d x}=2 \cos ^2 \frac{u}{2}+2 \sin \frac{u}{2} \cos \frac{u}{2}=2 \cos ^2 \frac{u}{2}\left(1+\tan \frac{u}{2}\right)\)
Separating the variables : \(\frac{d u}{2 \cos ^2(u / 2)[1+\tan (u / 2)]}=d x\)
⇒ \(\int \frac{\sec ^2(u / 2)}{2[1+\tan (u / 2)]} d u=\int d x+c \Rightarrow \log \left(1+\tan \frac{u}{2}\right)=x+c\)
∴ The General Solution is \(\log \left[1+\tan \left(\frac{x+y}{2}\right)\right]=x+c\)
Example. 10: solve : \(\frac{d y}{d x}-x \tan (y-x)=1\)
Solution:
Given equation is \(\frac{d y}{d x}-x \tan (\cdot y-x)=1\) …………………..(1)
Put \(y-x=z \Rightarrow \frac{d y}{d x}-1=\frac{d z}{d x} \Rightarrow \frac{d y}{d x}=\frac{d z}{d x}+1\) ……………………(2)
(1)and(2) => \(\frac{d z}{d x}+1-x \tan z=1 \Rightarrow \frac{d z}{d x}=x \tan z\)
Separating the variables: \(\frac{d z}{\tan z}=x d x \Rightarrow \int \cot z d z=\int x d x+c \Rightarrow \log |\sin z|=\frac{x^2}{2}+\frac{c}{2}\)
∴ The general solution of (1) is \(2 \log |\sin (y-x)|=x^2+c\)
Example. 11: Solve : \(\left(\frac{x+y+a}{x+y-b}\right) \frac{d y}{d x}=\frac{x+y-a}{x+y+b}\)
Solution:
Given equation is
\(\left(\frac{x+y+a}{x+y-b}\right) \frac{d y}{d x}=\frac{x+y-a}{x+y+b}\)Put x + y = z in the given equation => \(1+\frac{d y}{d x}=\frac{d z}{d x}\)
∴ Given differential equation becomes \(\left(\frac{z+a}{z-b}\right)\left(\frac{d z}{d x}-1\right)=\frac{z-a}{z+b}\)
⇒ \(\frac{d z}{d x}=1+\frac{(z-a)(z-b)}{(z+a)(z+b)}=\frac{z^2+z(a+b)+a b+z^2-z(a+b)+a b}{(z+a)(z+b)}\)
⇒ \(\frac{d z}{d x}=\frac{2\left(z^2+a b\right)}{z^2+z(a+b)+a b}\)
Separating the variables: \(\Rightarrow \frac{z^2+z(a+b)+a b}{z^2+a b} d z=2 d x\)
⇒ \(\int\left[\frac{z^2+a b}{z^2+a b}+\frac{z(a+b)}{z^2+a b}\right] d z=\int 2 d x+c\)
⇒ \(\int d z+\frac{a+b}{2} \int \frac{2 z}{z^2+a b} d z=2 x+c \Rightarrow z+\frac{a+b}{2} \log \left|z^2+a b\right|=2 x+c\)
∴ The general solution is \(x+y+\frac{a+b}{2} \log \left|(x+y)^2+a b\right|=2 x+c\)
Example. 12: Find the equation of the curve passing through the point (1,1) whose differential equation is (y- yx) dx + (x+xy) dy = 0.
Solution :
Given equation is (y- yx) dx + (x+xy) dy = 0 => y (1 – x) dx + x (1+ y) dy =0
Separating the variables: \(\frac{1-x}{x} d x+\frac{1+y}{y} d y=0 \Rightarrow\left(\frac{1}{x}-1\right) d x+\left(\frac{1}{y}+1\right) d y=0\)
⇒ \(\int\left(\frac{1}{x}-1\right) d x+\int\left(\frac{1}{y}+1\right) d y=c \Rightarrow \log |x|-x+\log |y|+y=c\)
Given the curve passes through the point (1,1): = log 1 -1+ log 1 +1 = c => c = 0
∴ The equation of the curve passing through the point (1,1) is \(\log |x y|+y-x=0 \Rightarrow x y=e^{x-y}\)