Differential Equations of First Order and First Degree Solved Example Problems Exercise 2.7
Example. 1 Solve \(\left(x^2+1\right) \frac{d y}{d x}+4 x y=\frac{1}{x^2+1}\)
Solution.
Given Equation
\(\left(x^2+1\right) \frac{d y}{d x}+4 x y=\frac{1}{x^2+1}\)Reducing the given equation to standard form, by dividing with \(\left(x^2+1\right)\)
⇒ \(\frac{d y}{d x}+\frac{4 x}{x^2+1} y=\frac{1}{\left(x^2+1\right)^2} \text { where } \mathrm{P}=\frac{4 x}{x^2+1}, \mathrm{Q}=\frac{1}{\left(x^2+1\right)^2}\) ……….(1)
⇒ \(\int \mathrm{P} d x=\int \frac{4 x}{x^2+1} d x=2 \log \left(x^2+1\right)=\log \left(x^2+1\right)^2\)
Then l.F = \(\exp \left[\int P d x\right]=\exp \left[\log \left(x^2+1\right)^2\right]=\left(x^2+1\right)^2\)
∴ G. S. of (1) is y (I.F.) = \(\int \mathrm{Q}(\mathrm{I} \cdot \mathrm{F}) d x+c\)
⇒ \(y\left(x^2+1\right)^2=\int \frac{1}{\left(x^2+1\right)^2} \cdot\left(x^2+1\right)^2 d x=\int d x=x+c\)
Differential Equations Of First Order And First Degree Exercise 2.7
Example. 2: Solve \(x \frac{d y}{d x}+2 y-x^2 \log x=0\)
Solution:
Given Equation
\(x \frac{d y}{d x}+2 y-x^2 \log x=0\)Divide the equation with x to reduce it to standard form :
⇒ \(\frac{d y}{d x}+\frac{2}{x} y=x \log x\) ……………….(1)
where p = 2/x and Q = x log x
⇒ \(\int \mathrm{P} d x=\int \frac{2}{x} d x=2 \log x=\log x^2 \text { then I.F. }=e^{\log x^2}=x^2\)
The G, S. of (1)is y(I.F.) = \(\int \text { Q (I.F.) } d x+c \Rightarrow y\left(x^2\right)=\int(x \log |x|) x^2 d x+c\)
⇒ \(x^2 y=\int x^3 \log |x| d x+c=\frac{x^4}{4} \log x-\int \frac{x^4}{4} \cdot \frac{1}{x} d x+c\)
⇒ \(x^2 y=\frac{x^4}{4} \log |x|-\int \frac{x^3}{4} d x+c \Rightarrow x^2 y=\frac{x^4}{4} \log |x|-\frac{x^4}{16}+c\)
Homogeneous Equations Solved Problems Exercise 2.7
Example. 3: Solve \(x \cos x \frac{d y}{d x}+(x \sin x+\cos x) y=1\)
Solution.
Given Equation
\(x \cos x \frac{d y}{d x}+(x \sin x+\cos x) y=1\)Dividing the given equation by x cos x, we get
⇒ \(\frac{d y}{d x}+\frac{x \sin x+\cos x}{x \cos x} y=\frac{1}{x \cos x}\)
where \(\mathrm{P}=\frac{x \sin x+\cos x}{x \cos x}, \mathrm{Q}=\frac{1}{x \cos x}\)
⇒ \(\int \mathrm{P} d x=\int \frac{x \sin x+\cos x}{x \cos x} d x=\int\left(\tan x+\frac{1}{x}\right) d x\)
= \(\int \tan x d x+\int \frac{1}{x} d x=\log |\sec x|+\log |x|=\log |x \sec x|\)
∴ \(\text { I.F. }=\exp \left[\int \mathrm{P} d x\right]=\exp (\log |x \sec x|)=x \sec x\)
The G. S. of (1) is y(l.F.) = \(\int \mathrm{Q}(\mathrm{I} . \mathrm{F}) d x+c\)
y \((x \sec x)=\int \frac{1}{x \cos x}(x \sec x) d x+c \Rightarrow y(x \sec x)=\int \sec ^2 x d x+c \Rightarrow x y \sec x=\tan x+c\)
Methods To Find Integrating Factors For Exercise 2.7
Example. 4: Solve \(x(x-1) \frac{d y}{d x}-(x-2) y=x^3(2 x-1)\)
Solution.
Given Equation
\(x(x-1) \frac{d y}{d x}-(x-2) y=x^3(2 x-1)\)Dividing the given equation with x(x -1), we get
⇒ \(\frac{d y}{d x}-\frac{x-2}{x(x-1)} y=\frac{x^2(2 x-1)}{x-1} \text { where } \mathrm{P}=-\frac{x-2}{x(x-1)}, \mathrm{Q}=\frac{x^2(2 x-1)}{x-1}\) ………….(1)
⇒ \(\int \mathrm{P} d x=\int-\frac{x-2}{x(x-1)} d x=\int\left(\frac{1}{x-1}-\frac{2}{x}\right) d x=\log (x-1)-2 \log x\)
∴ \(\text { I.F. }=\exp \left[\int \mathrm{P} d x\right]=e^{\log (x-1)} \cdot e^{-2 \log x}\)
= \(e^{\log (x-1)} \cdot e^{\log x^{-2}}=e^{\log (x-1) x^{-2}}=(x-1) x^{-2}=\frac{x-1}{x^2}\)
∴ G. S. of (1) is y (I.F.) = \(\int \mathrm{Q}(\text { I.F.) } d x+c\)
⇒ \(\frac{y(x-1)}{x^2}=\int \frac{x^2(2 x-1)}{x-1} \cdot \frac{x-1}{x^2} d x+c=\int(2 x-1) d x+c\)
⇒ \(\frac{y(x-1)}{x^2}=\int \frac{2 x^2}{2} d x-\int d x+c=x^2-x+c\)
Solutions For Exercise 2.7 First-Order Homogeneous Equations
Example. 5: Solve \(\left(y-e^{\sin ^{-1} x}\right) \frac{d x}{d y}+\sqrt{1-x^2}=0, \quad|x|<1\)
Solution:
Given Equation
\(\left(y-e^{\sin ^{-1} x}\right) \frac{d x}{d y}+\sqrt{1-x^2}=0, \quad|x|<1\)Gien equation can be written as: \(\sqrt{1-x^2} \frac{d y}{d x}+y=e^{\text{Sin}^{-1}} x \Rightarrow \frac{d y}{d x}+\frac{1}{\sqrt{1-x^2}} y\)
= \(\frac{e^{\sin ^{-1} x}}{\sqrt{1-x^2}} \ldots \ldots \ldots\)
where \(\left.\mathrm{P}=\frac{1}{\sqrt{1-x^2}}, \mathrm{Q}=\frac{e^{\sin ^{-1} x}}{\sqrt{1-x^2}}\)
(Note.|x|<1 \(\Rightarrow\left(1-x^2\right)>0\right)\)
I.F. = \(\exp \left(\int \mathrm{P} d x\right)=\exp \left(\int \frac{d x}{\sqrt{1-x^2}}\right)\)
= \(\exp \left(\text{Sin}^{-1} x\right)=e^{\text{Sin}^{-1} x}\)
where \(\mathrm{P}=\frac{1}{\sqrt{1-x^2}}, \mathrm{Q}=\frac{e^{\sin ^{-1} x}}{\sqrt{1-x^2}}\)
(Note. |x|<1 ⇒ \(\left(1-x^2\right)>0\))
General solution of (1) is y (I.F.) = \(\int \mathrm{Q}\) (I.F.) dx+c
⇒ y e^{\text{Sin}^{-1} x}=\int \frac{e^{\text{Sin}^{-1} x}}{\sqrt{1-x^2}} \cdot e^{\text{Sin}^{-1} x} d x+c=\int \frac{e^{2 \text{Sin}^{-1} x}}{\sqrt{1-x^2}} d x+c[/latex]
⇒ \(y e^{\sin ^{-1} x}=\int e^{2 t} d t+c\) where t = \(\sin ^{-1} x \Rightarrow d t=\frac{d x}{\sqrt{1-x^2}}\)
⇒ \(y e^{\text{Sin}^{-1} x}=\frac{e^{2 t}}{2}+c=\frac{e^{2 \text{Sin}^{-1} x}}{2}+c \Rightarrow 2 y e^{\text{Sin}^{-1} x}=e^{2 \text{Sin}^{-1} x}+c\)
Step-By-Step Solutions For Exercise 2.7 Differential Equations
Example 6. Solve \(\frac{d y}{d x}+\frac{y}{(1-x) \sqrt{x}}=1-\sqrt{x}\)
Solution.
Given linear equation is \(\frac{d y}{d x}+\frac{y}{(1-x) \sqrt{x}}=1-\sqrt{x}\) where P = \(\frac{1}{(1-x) \sqrt{x}}\)
⇒ \(\int \mathrm{P} d x=\int \frac{1}{(1-x) \sqrt{x}} d x\) . Put [katex]\sqrt{x}=t \Rightarrow \frac{1}{2 \sqrt{x}} d x=d t \Rightarrow \frac{1}{\sqrt{x}} d x=2 d t[/latex]
= \(2 \int \frac{d t}{1-t^2}=2 \frac{1}{2} \log \frac{1+t}{1-t}=\log \frac{1+\sqrt{x}}{1-\sqrt{x}}\)
∴ \(\text { I.F. }=\exp \log \frac{1+\sqrt{x}}{1+\sqrt{x}}=\frac{1+\sqrt{x}}{1-\sqrt{x}}\)
∴ (\(\exp (\log x)=e^{\log x}=x\))
∴ G.S of given equation is \(y\left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right)=\int(1-\sqrt{x})\left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right) d x+c=\int(1+\sqrt{x}) d x+c\)
⇒ \(y\left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right)=x+\frac{x^{3 / 2}}{3 / 2}+c \Rightarrow y\left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right)=x+\frac{2}{3} x^{3 / 2}+c\)
Methods For Solving Exercise 2.7 Differential Equations
Example.7. Solve \(y d x-x d y+\log x d x=0 \text { (or) Solve } x \frac{d y}{d x}-y=\log x\)
Solution.
Given equation is \(y d x-x d y+\log x d x=0\)
⇒ \(\frac{d y}{d x}-\frac{1}{x} y=\frac{\log x}{x}\) This is a linear equation in y.
Where \(\mathrm{P}=-\frac{1}{x}, \mathrm{Q}=(\log x) / x . \text { I.F. }=e^{\int(-1 / x) d x}=e^{-\log x}=e^{\log x^{-1}}=1 / x\)
The GS. of (1) is \(y(1 / x)=\int \frac{1}{x} \cdot \frac{\log x}{x} d x+c=\int \frac{1}{x^2} \log x d x+c\)
= \(-\frac{1}{x} \log x-\int\left(-\frac{1}{x}\right) \frac{1}{x} d x+c \text { (Integrating by parts) }\)
= \(-\frac{1}{x} \log x+\int \frac{1}{x^2} d x+c=-\frac{1}{x} \log x-\frac{1}{x}+c\)
∴ G.S. is \(y=-\log x-1+c x \Rightarrow y=c x-(1+\log x)\)
Aliter: G. E. can be written as (x dy – y dx) – log x = 0
⇒ \(\frac{x d y-y d x}{x^2}-\frac{1}{x^2} \log x=0 \Rightarrow d\left(\frac{y}{x}\right)+(\log x) d\left(\frac{1}{x}\right)=0\)
⇒ \(\int d\left(\frac{y}{x}\right)+\int \log x d\left(\frac{1}{x}\right)=c \Rightarrow\left(\frac{y}{x}\right)+\left(\frac{1}{x}\right) \log x-\int\left(\frac{1}{x^2}\right) d x \Rightarrow\left(\frac{y}{x}\right)+\left(\frac{1}{x}\right) \log x+\left(\frac{1}{x}\right)=c\)
Solution is \((y / x)+(1 / x)(1+\log x)=c\)
Exercise 2.7 Solutions For First-Order Differential Equations
Example. 8: Obtain the equation of the curve satisfying the differential equation \(\left(1+x^2\right) \frac{d y}{d x}+2 x y-4 x^2=0\) and passing through the origin
Solution:
Given equation is \(\left(1+x^2\right) \frac{d y}{d x}+2 x y=4 x^2\) ……..(1)
Dividing (1) by \(\left(1+x^2\right)\) to reduce it to standard form
⇒ \(\frac{d y}{d x}+\frac{2 x}{1+x^2} y=\frac{4 x^2}{1+x^2} \text { where } \mathrm{P}=\frac{2 x}{1+x^2}, \mathrm{Q}=\frac{4 x^2}{1+x^2}\) …………(2)
Now, \(\int \mathrm{P} d x=\int \frac{2 x}{1+x^2} d x=\log \left(1+x^2\right)\)
∴ \(\text { I.F. }=\exp \left(\int \mathrm{P} d x\right)=\exp \left[\log \left(1+x^2\right)\right]=1+x^2\)
∴ G.S. of (1) is y(I.F.) = \(\int \mathrm{Q}(\mathrm{I} \cdot \mathrm{F}) d x+c \Rightarrow y\left(1+x^2\right)=\int \frac{4 x^2}{1+x^2} \cdot\left(1+x^2\right) d x+c=\int 4 x^2 d x+c\)
⇒ \(y\left(1+x^2\right)=\left(4 x^3 / 3\right)+c\)
Given the curve passes through the origin (0,0) => 0 = 0 +c => c = 0
∴ the equation if the required curve is \(3 y\left(1+x^2\right)=4 x^3\)