Differential Equations of First Order and First Degree Solved Example Problems Exercise 2
Example. 1: Solve \(\left(e^y+1\right) \cos x d x+e^y \sin x d y=0\)
Solution.
Given, equation is: \(\left(e^y+1\right) \cos x d x+e^y \sin x d y=0\) …………………(1)
(1) is in the form Mdx + Ndy = 0 where \(\mathrm{M}=\left(e^y+1\right) \cos x\) and \(\mathrm{N}=e^y \sin x\) Now \(\frac{\partial \mathrm{M}}{\partial y}=e^y \cos x, \frac{\partial \mathrm{N}}{\partial x}=e^y \cos x \Rightarrow \frac{\partial \mathrm{M}}{\partial y}=\frac{\partial \mathrm{N}}{\partial x}\) =>(1) is an exact equation.
⇒ \(\int^x \mathrm{M} d x=\int^x\left(e^y+1\right) \cos x d x=\left(e^y+1\right) \sin x\) ……………………..(2)
(Integrating M w.r.t. x, treating y as constant.)
⇒ \(\int\)(terms of N not containing x) dy=\(\int 0 . d y=0\) …………………….(3)
(2) + (3)=> The general solution of (1) is \(\left(e^y+1\right) \sin x+0=c \Rightarrow\left(e^y+1\right) \sin x=c\)
Example. 2: Solve: \((2 x y+y-\tan y) d x+\left(x^2-x \tan ^2 y+\sec ^2 y\right) d y=0\)
Solution.
Given equation is \((2 x y+y-\tan y) d x+\left(x^2-x \tan ^2 y+\sec ^2 y\right) d y=0\) ………………….. (1)
(1) is in the form Max + Ndy=0 where M = 2xy + y-tan y and \(\mathrm{N}=x^2-x \tan ^2 y+\sec ^2 y\)
Now \(\frac{\partial \mathrm{M}}{\partial y}=2 x+1-\sec ^2 y=2 x-\left(\sec ^2 y-1\right)=2 x-\tan ^2 y\)
⇒ \(\frac{\partial \mathrm{N}}{\partial x}=2 x-\tan ^2 y \Rightarrow \frac{\partial \mathrm{M}}{\partial y}=\frac{\partial \mathrm{N}}{\partial x}\) => (1) is an exact equation.
(1) Integrating M w.r.t. x, treating y as constant.
⇒ \(\int^x \mathrm{M} d x=\int^x(2 x y+y-\tan y) d x=\frac{2 x^2}{2} y+(y-\tan y) x=x^2 y+(y-\tan y) x\) ……………………(2)
(2) \(\int\)(terms of N not containing x) dy = \(\int \sec ^2 y d y\) =tan y …………………….(3)
The general solution of (1) is (2) + (3) = \(c \Rightarrow x^2 y+(y-\tan y) x+\tan y=c\)
Example. 3. Show that the equation \(\left(y^2 e^{x y^2}+4 x^3\right) d x+\left(2 x y e^{x y^2}-3 y^2\right) d y=0\) is exact and hence solve it.
Solution.
Given equation is \(\left(y^2 e^{x y^2}+4 x^3\right) d x+\left(2 x y e^{x y^2}-3 y^2\right) d y=0\) ……………………(1)
(1) is in the form Mdx +Ndy=0 where \(\mathrm{M}=y^2 e^{x y^2}+4 x^3 \text { and } \mathrm{N}=2 x y e^{x y^2}-3 y^2\)
Now \(\frac{\partial \mathrm{M}}{\partial y}=2 y e^{x y^2}+y^2\left(e^{x y^2}\right)(2 x y)=2 y e^{x y^2}+2 x y^3 e^{x y^2}\)
and \(\frac{\partial \mathrm{N}}{\partial x}=2 y e^{x y^2}+2 x y \cdot e^{x y^2}\left(y^2\right)=2 y e^{x y^2}+2 x y^3 e^{x y^2}\)
⇒ \(\frac{\partial \mathrm{M}}{\partial y}=\frac{\partial \mathrm{N}}{\partial x}\) => (1) is an exact equation.
(1) Integrating M w.r.t. x treating y as constant.
⇒ \(\int^x \mathrm{M} d x=\int\left(y^2 e^{x y^2}+4 x^3\right) d x=[latex]y^2 \frac{e^{x y^2}}{x^2}\)+\(4 \frac{x^4}{4}=e^{x y^2}+x^4\) …………………..(2)
(2) \(\int\)( terms of N not containing x) dy = \(\int\left(-3 y^2\right) d y=-3 \cdot \frac{y^3}{3}=-y^3\) ………………………(3)
The general solution of (1) is (2) + (3) = c => \(e^{x y^2}+x^4-y^3=c\)
Example. 4. Show that the equation \(x d x+y d y=\frac{a^2(x d y-y d x)}{x^2+y^2}\) is exact and solve it.
Solution:
Given x dx+y dy = \(\frac{a^2(x d y-y d x)}{x^2+y^2} \Rightarrow\left(x+\frac{a^2 y}{x^2+y^2}\right) d x+\left(y-\frac{a^2 x}{x^2+y^2}\right) d y=0\)……(1)
is in the form M dx + N dy = 0 where M = \(x+\frac{a^2 y}{x^2+y^2}\) and \(\mathrm{N}=y-\frac{a^2 x}{x^2+y^2}\)
⇒ \(\frac{\partial \mathrm{M}}{\partial y}=a^2\left[\frac{\left(x^2+y^2\right) \cdot 1-y(2 y)}{\left(x^2+y^2\right)^2}\right]\)
= \(\frac{a^2\left(x^2-y^2\right)}{\left(x^2+y^2\right)^2}\)
And \(\frac{\partial \mathrm{N}}{\partial x}=-a^2\left[\frac{\left(x^2+y^2\right) \cdot 1-x(2 x)}{\left(x^2+y^2\right)^2}\right]\)
= \(\frac{-a^2\left(y^2-x^2\right)}{\left(x^2+y^2\right)}\)
= \(\frac{a^2\left(x^2-y^2\right)}{\left(x^2+y^2\right)^2} \Rightarrow \frac{\partial \mathrm{M}}{\partial y}=\frac{\partial \mathrm{N}}{\partial x}\)
∴ (1) is an exact equation.
(1) Integrating M w.r.t. x, treating y as constant. \(\int^x M d x=\int\left(x+\frac{a^2 y}{x^2+y^2}\right) d x\)
= \(\frac{x^2}{2}+a^2 y \cdot \frac{1}{y} \text{Tan}^{-1}\left(\frac{x}{y}\right)=\frac{x^2}{2}+a^2 \text{Tan}^{-1}\left(\frac{x}{y}\right)\)………..(2)
(2) \(\int\) (terms of N not containing x) dy = \(\int y d y=\frac{y^2}{2}\)
∴ The general solution of (1) is (2) +(3)=c
⇒ \(\frac{x^2}{2}+a^2 \text{Tan}^{-1}\left(\frac{x}{y}\right)+\frac{y^2}{2}=c \Rightarrow x^2+y^2+2 a^2 \text{Tan}^{-1}\left(\frac{x}{y}\right)=2 c\)
Example. 5: Solve: \(\left[y\left(1+\frac{1}{x}\right)+\cos y\right] d x+(x+\log x-x \sin y) d y=0\)
Solution.
Given equation is \(\left[y\left(1+\frac{1}{x}\right)+\cos y\right] d x+(x+\log x-x \sin y) d y=0\) …………….(1)
(1) is in the form Mdx + Ndy = 0 where \(\mathrm{M}=y\left(1+\frac{1}{x}\right)+\cos y \text { and } \mathrm{N}=x+\log x-x \sin y\)
⇒ \(\frac{\partial \mathrm{M}}{\partial y}=1+\frac{1}{x}-\sin y\) and
⇒ \(\frac{\partial \mathrm{N}}{\partial x}=1+\frac{1}{x}-\sin y\)
⇒ \(\frac{\partial \mathrm{M}}{\partial y}=\frac{\partial \mathrm{N}}{\partial x}\)
∴ (1) is an exact equation.
Now (1) Integrating M w.r.t. x, treating y as constant \(\int^x \mathrm{M} d x\)
= \(\int^x\left[y\left(1+\frac{1}{x}\right)+\cos y\right] d x=y(x+\log |x|)+x \cos y\) ……………………(2)
(2) \(\int\)(terms of N not involving x) dx = \(\int 0 \cdot d x=0\) …………….(3)
∴ The general solution of (1) is (2)+(3) = c
⇒ \(y(x+\log |x|)+x \cos y+0=c \Rightarrow y(x+\log |x|)+x \cos y=c\)
Example. 6. Solve \(\frac{d y}{d x}+\frac{a x+h y+g}{h x+b y+f}=0\) and show that this differential equation represents a family of conics.
Solution.
Given equation is \(\frac{d y}{d x}+\frac{a x+h y+g}{h x+b y+f}=0 \Rightarrow(a x+h y+g) d x+(h x+b y+f) d y=0\) ….(1)
(1) is of the form Mdx + Ndy = 0 where M -ax + hy + g and N = hx + by + f
⇒ \(\frac{\partial \mathrm{M}}{\partial y}=h\)
and \(\frac{\partial \mathrm{N}}{\partial y}\)=h
⇒ \(\frac{\partial \mathrm{M}}{\partial y}=\frac{\partial \mathrm{N}}{\partial x}\)
∴ (1) is an exact equation.
Now (1) Integrating M w.r.t. x, treating y as constant. \(\int^x \mathrm{M} d x=\int(a x+h y+g) d x=\frac{a x^2}{2}+h x y+g x\)
(2)\(\int\)(terms of N free from x)dy = \((b y+f) d y=\frac{b y^2}{2}+f y\)
∴ The general solution of (1) is (2) + (3) = c.
⇒ \(\frac{a x^2}{2}+h y x+g x+\frac{b y^2}{2}+f y=c\)
⇒ \(a x^2+2 h x y+b y^2+2 g x+2 f y+c=0\) where c = -2c
Clearly, this equation represents a family of conics.
Example. 7: Solve (y cos 2 x – 2 x ) dx + (sin x + cos y) dy = 0, it being given that y = 0 when x =0
Solution.
Given equation is (y cos 2 x – 2 x) dx + (sin x + cos y) dy = 0 ……………….. (1)
(1) is of the form Mdx + Ndy = 0 where M = y cos x – 2x and N = sin x + cos y
⇒ \(\frac{\partial \mathrm{M}}{\partial y}=\cos x, \frac{\partial \mathrm{N}}{\partial x}\)=cos x
⇒ \(\frac{\partial \mathrm{M}}{\partial y}=\frac{\partial \mathrm{N}}{\partial x}\)
∴ (1) is an exact equation.
Now (1) Integrating M w.r.t. x, treating y as constant. \(\int^x \mathrm{M} d x=\int^x(y \cos x-2 x) d x=y \sin x-x^2\) ……………….(2)
(2) \(\int\)(terms of N free from x)dy = \(\int \cos y d y=\sin y\) ……………………….(3)
∴ The general solution of (1) is (2) + (3) = c => \(y \sin x-x^2+\sin y=c\) ……………………… (4)
Given y = 0 when x = 0. Substituting these values in (4)
0sin 0 – 0 + sin0 = c => 0 – 0 + 0 = c => c = 0
The required solution is \(y \sin x-x^2+\sin y=0\)
Example. 8: Solve (r + sinθ-cosθ)dr + r(sinθ + cosθ)dθ = O
Solution.
Given equation is (r + sinθ-cosθ)dr + r(sinθ + cosθ)dθ = 0 …………………. (1)
(1) is of the form Mdθ + Ndr where M = r(sin θ + cos θ) and N = r + sin θ – cos θ
⇒ \(\frac{\partial \mathrm{M}}{\partial r}=\sin \theta+\cos \theta, \frac{\partial \mathrm{N}}{\partial \theta}=\cos \theta+\sin \theta\)
⇒ \(\frac{\partial \mathrm{M}}{\partial r}=\frac{\partial \mathrm{N}}{\partial \theta}\)
∴ (1) is an exact equation.
(1) Integrating M w.r.t. θ, treating r as constant \(\int^\theta M d \theta=\int r(\sin \theta+\cos \theta) d \theta=r(-\cos \theta+\sin \theta)\) ……………………..(2)
(2) \(\)(terms of N not containing θ) dr = \(\int r d r=\frac{r^2}{2}\)………………………(3)
The genera] solution of (1) is (2) + (3) = \(\frac{c}{2}\)
⇒ \(r(\sin \theta-\cos \theta)+\frac{r^2}{2}=\frac{c}{2} \Rightarrow r^2+2 r(\sin \theta-\cos \theta)=c\)