Differential Equations of First Order and First Degree Solved Example Problems Exercise 2.10
Example. 1: Solve \((x+1) \frac{d y}{d x}+1=e^{x-y}\). Also, find the solution for which y(0) = 0
Solution.
Given equation is \((x+1) \frac{d y}{d x}+1=e^{x-y}\) ….(1)
⇒ \(\frac{d y}{d x}+\frac{1}{x+1}=\frac{e^x \cdot e^{-y}}{x+1}\)
Multiplying with \(e^y\), we get: \(e^y \frac{d y}{d x}+\frac{1}{x+1} e^y=\frac{e^x}{x+1}\) = which is Bernoulli’s equation… (2)
Let \(e^y=u \Rightarrow e^y \frac{d y}{d x}=\frac{d u}{d x}\) ……………..(3)
(2) and (3) \(\Rightarrow \frac{d u}{d x}+\frac{1}{x+1} u=\frac{e^x}{x+1}\) which is linear in u and x …………………………(4)
where \(\mathrm{P}=\frac{1}{x+1} \text { and } \mathrm{Q}=\frac{e^x}{x+1}\)
Now \(\text { I.F. }=\exp \left(\int \frac{1}{x+1} d x\right)=e^{\log (x+1)}=x+1\)
The G.S of (4) is \(u(x+1)=\int \frac{e^x}{x+1}(x+1) d x+c=\int e^x d x+c \Rightarrow u(x+1)=e^x+c\)
∴ The general solution of (1) is \(e^y(x+1)=e^x+c\)
y(0) = 0=> value of y at (x = 0) is 0.
∴ \(e^0(0+1)=e^0+c \Rightarrow 1=1+c \Rightarrow c=0\)
The required solution when \(y(0)=0 \text { is } e^y(x+1)=e^x\)
Example. 2: Solve \(\frac{d y}{d x}+\left(2 x \text{Tan}^{-1} y-x^3\right)\left(1+y^2\right)\)=0
Solution.
Given equation is \(\frac{d y}{d x}+\left(2 x \text{Tan}^{-1} y-x^3\right)\left(1+y^2\right)=0\)………(1)
Dividing (1) by \(\left(1+y^2\right)\) and rearranging the equation:
⇒ \(\frac{1}{1+y^2} \frac{d y}{d x}+2 x \mathrm{Tan}^{-1} y=x^3\) which is Bernoulli’s equation……(2)
Let \(\text{Tan}^{-1} y=u \Rightarrow \frac{1}{1+y^2} \frac{d y}{d x}=\frac{d u}{d x}\)……….(3)
(2) and (3) ⇒ \(\frac{d u}{d x}+2 x u=x^3\) which is linear … (4) in u and x where \(\mathrm{P}=2 x\) and \(\mathrm{Q}=x^3\)
∴ I.F. \(=e^{\int 2 x d x}=e^{x^2}\).
Now the G. S. of (4) is \(u. e^{x^2}=\int x^3 e^{x^2} d x+c=\frac{1}{2} \int e^{t^t} \cdot t d t+c\)
where \(t=x^2 \Rightarrow u e^{x^2}=\frac{1}{2}\left(t e^t-e^t\right)+c\)
Substituting \(u=\text{Tan}^{-1} \quad y\) and \(t=x^2\)
The general solution of (1) is \(2 e^{x^2} \text{Tan}^{-1} y=\left(x^2-1\right) e^{x^2}+2 c\)
Example. 3: Solve \(\frac{d z}{d x}+\frac{z}{x} \log z=\frac{z}{x^2}(\log z)^2, z>0 \text { and } x>0\)
Solution.
Given \(\frac{d z}{d x}+\frac{z}{x} \log z=\frac{z}{x^2}(\log z)^2\) …………..(1)
(1) is Bernoulli’s equation. Then multiply (1) by \(z^{-1}(\log z)^{-2}\):
⇒ \(z^{-1}(\log z)^{-2} \frac{d z}{d x}+\frac{(\log z)^{-1}}{x}=\frac{1}{x^2}\) ………………(2)
Let \((\log z)^{-1}=u \Rightarrow z^{-1}(\log z)^{-2} \frac{d z}{d x}=-\frac{d u}{d x}\) ……………..(3)
(2) and (3) ⇒ \(-\frac{d u}{d x}+\frac{u}{x}=\frac{1}{x^2} \Rightarrow \frac{d u}{d x}-\frac{1}{x} u=-\frac{1}{x^2}\) ……………..(4)
(4) is a linear equation in u and x where \(\mathrm{P}=-\frac{1}{x}, \mathrm{Q}=-\frac{1}{x^2}\)
Now \(\text { I.F. }=\exp \left(\int \mathrm{P} d x\right)=\exp \left(-\frac{1}{x} d x\right)=e^{-\log x}=e^{\log x^{-1}}=\frac{1}{x}\)
The general solution of (4) is \(u\left(\frac{1}{x}\right)=\int\left(-\frac{1}{x^2}\right) \frac{1}{x} d x+c \Rightarrow u\left(\frac{1}{x}\right)=\int-x^{-3} d x+c=\frac{1}{2 x^2}+c\) ………..(5)
substitution \(u=(\log z)^{-1}\) in (5), the general solution of (1) is \(\frac{1}{x \log z}=\frac{1}{2 x^2}+c \Rightarrow 2 x=\log z+2 x^2 c \log z\)
Example. 4. Solve \(\frac{x d x+y d y}{x d y-y d x}=\sqrt{\frac{a^2-x^2-y^2}{x^2+y^2}}\)
Solution:
Given
\(\frac{x d x+y d y}{x d y-y d x}=\sqrt{\frac{a^2-x^2-y^2}{x^2+y^2}}\)Let \(x=r \cos \theta\) and \(y=r \sin \theta\).
Then \(r^2=x^2+y^2, \theta=\text{Tan}^{-1}(y / x)\)
dx = \(d r \cdot \cos \theta-r \sin \theta d \theta, d y=r \cos \theta \cdot d \theta+d r \cdot \sin \theta\)
Then (1) becomes: \(\frac{r \cos \theta(\cos \theta \cdot d r-r \sin \theta d \theta)+r \sin \theta(r \cos \theta d \theta+d r \cdot \sin \theta)}{r \cos \theta(r \cos \theta d \theta+d r \cdot \sin \theta)-r \sin \theta(d r \cdot \cos \theta-r \sin \theta d \theta)}\)
= \(\sqrt{\frac{a^2-r^2}{r^2}} \Rightarrow \frac{\left(r \cos ^2 \theta+r \sin ^2 \theta\right) d r}{\left(r^2 \cos ^2 \theta+r^2 \sin ^2 \theta\right) d r}\)
= \(\sqrt{\frac{a^2-r^2}{r^2}} \Rightarrow \frac{1}{r} \frac{d r}{d \theta}=\frac{\sqrt{a^2-r^2}}{r}\)
⇒ \(\frac{d r}{\sqrt{a^2-r^2}}=d \theta \Rightarrow \sin ^{-1} \frac{r}{a}=\theta+c \Rightarrow r=a \sin (\theta+c)\)
The G.S. of (1) is \(\sqrt{x^2+y^2}=a \sin \left[\text{Tan}^{-1}\left(\frac{y}{x}\right)+c\right]\)