Differential Equations of First Order and First Degree Solved Example Problems Exercise 2.8
Example. 1: Solve \(\left(1+y^2\right) d x=\left(\text{Tan}^{-1} y-x\right) d y\)
Solution.
Given equation
\(\left(1+y^2\right) d x=\left(\text{Tan}^{-1} y-x\right) d y\)Given equation can be written as \(\left(1+y^2\right) \frac{d x}{d y}+x=\text{Tan}^{-1} y\)
Dividing by \(\left(1+y^2\right)\) to reduce this to standard form: [/latex]\frac{d x}{d y}+\frac{1}{1+y^2} x[/latex] = \(\frac{\text{Tan}^{-1} y}{1+y^2}\)……(1)
Where \(P_1=\frac{1}{1+y^2}\) and \(Q_1=\frac{\text{Tan}^{-1} y}{1+y^2}\) are functions of y alone.
Now \(\int P_1 d y=\frac{1}{1+y^2} d y=\text{Tan}^{-1} y\)
∴ I.F. = \(\exp \left(\int \mathrm{P} d y\right)=e^{\text{Tan}^{-1} y}\)
∴ G. S. of (1) is x (I.F.) \(=\int \mathrm{Q}_1\)
(I.F. ) dy+\(c \Rightarrow x e^{\mathrm{Tan}^{-1} y}=\int \frac{\text{Tan}^{-1} y}{1+y^2} \cdot e^{\mathrm{Tan}^{-1} y} d y+c\)
Put \(\text{Tan}^{-1} y=u \Rightarrow \frac{d y}{1+y^2}=d u\)
∴ G. S.is \(x e^{\text{Tan}^{-1} y}=\int u e^u d u+c=u e^u-e^u+c\)
∴ \(x e^{\text{Tan}^{-1} y}=e^{\text{Tan}^{-1} y}\left(\text{Tan}^{-1} y-1\right)+c\)
Differential Equations Of First Order And First Degree Exercise 2.8
Example. 2: Solve \(\left(1+x+x y^2\right) \frac{d y}{d x}+\left(y+y^3\right)=0\)
Solution.
Given equation
\(\left(1+x+x y^2\right) \frac{d y}{d x}+\left(y+y^3\right)=0\)Given equation can be written as \(y\left(1+y^2\right) \frac{d x}{d y}+1+x\left(1+y^2\right)=0\)
⇒ \(\frac{d x}{d y}+\frac{x\left(1+y^2\right)}{y\left(1+y^2\right)}=\frac{-1}{y\left(1+y^2\right)} \Rightarrow \frac{d x}{d y}+\frac{1}{y} x=\frac{-1}{y\left(1+y^2\right)}\)
which is a linear equation in x
where \(\mathrm{P}_1=\frac{1}{y}\) and \(\mathrm{Q}_1=\frac{-1}{y\left(1+y^2\right)}\)
⇒ \(\int \mathrm{P}_1 d y=\int \frac{1}{y} d y=\log y\)
∴ I.F. = \(\exp \left(\int \mathrm{P}_1 d y\right)=e^{\log y}=y\)
∴ G. S. of (1) is x (I.F.) = \(\int \mathrm{Q}_1\) (I.F.) dy+c
⇒ x(y) = \(\int \frac{-1}{y\left(1+y^2\right)} \cdot y d y+c \Rightarrow x y=-\int \frac{1}{1+y^2} d y+c\)
⇒ xy = \(-\tan{Tan}^{-1} y+c \Rightarrow x y+\text{Tan}^{-1} y=c\)
Homogeneous Equations Solved Problems Exercise 2.8
Example. 3: Solve \(\left(x+2 y^3\right) \frac{d y}{d x}=y\)
Solution.
Given equation
\(\left(x+2 y^3\right) \frac{d y}{d x}=y\)Given equation may be written as \(y \frac{d x}{d y}=x+2 y^3 \Rightarrow \frac{d x}{d y}-\frac{1}{y} x=2 y^2\) ………(1)
where \(\mathrm{P}_1=-\frac{1}{y} \text { and } \mathrm{Q}_1=2 y^2\)
Now, \(\int \mathrm{P}_1 d y=\int\left(-\frac{1}{\dot{y}}\right) d y=-\log y \Rightarrow \text { I.F. }=\exp \left(\int \mathrm{P}_1 d y\right)=e^{-\log y}=\frac{1}{y}\)
Hence G. S. of (1) is x (I.F) \(=\int \mathrm{Q}_1(\mathrm{I} . \mathrm{F}) d y+c\)
⇒ \(x(1 / y)=\int 2 y^2(1 / y) d y+c=2\left(y^2 / 2\right)+c=y^2+c\)
∴ \(\text { G. S. of (1) is } x=y^3+c y\)
Methods To Find Integrating Factors For Exercise 2.8
Example 4. Solve \((x+y+1) \frac{d y}{d x}=1\)
Solution. Given \((x+y+1) \frac{d y}{d x}=1 \Rightarrow \frac{d x}{d y}=x+y+1 \Rightarrow \frac{d x}{d y}-x=(y+1)\)
This is a linear equation in x. I.F. = \(e^{\int-d y}=e^{-y}\)
∴ The GS. is \(x e^{-y}=\int(y+1) e^{-y} d y+c=\int y e^{-y} d y+\int e^{-y} d y+c\)
⇒\(x e^{-y}=y\left(-e^{-y}\right)-\int\left(-e^{-y}\right) d y-e^{-y}+c=-y e^{-y}+\int e^{-y} d y-e^{-y}+c\)
⇒ \(x e^{-y}=-y e^{-y}-e^{-y}-e^{-y}+c \Rightarrow x+y+2=c e^y\)