(B) : Given differential equation- \(\left(1+y^{2}\right) d x=\left(\tan ^{-1} y-x\right) d y\) Can be written as \(\frac{d y}{d x}=\frac{\tan ^{-1} y-x}{1+y^{2}}\) Or \(\quad \frac{d y}{d x}+\frac{1}{1+y^{2}} \cdot x=\frac{\tan ^{-1} y}{1+y^{2}}\) It is a linear differential of the form \(\frac{\mathrm{dx}}{\mathrm{dy}}+\mathrm{Px}=\mathrm{Q}\) Where \(\quad P=\frac{1}{1+y^{2}}, \quad Q=\frac{\tan ^{-1} y}{1+y^{2}}\) \(\therefore\) Integrating factor (IF) \(=\mathrm{e}^{\mathrm{Pdy}}\) \(=\mathrm{e}^{\int \frac{1}{1+\mathrm{y}^{2}} \mathrm{dy}}=\mathrm{e}^{\tan ^{-1} \mathrm{y}}\) \(\mathrm{x} . \mathrm{IF}=\int \mathrm{Q} \cdot \mathrm{IF}+\mathrm{C}\) So, solution is \(\mathrm{e}^{\tan ^{-1} y} \cdot x=\int \frac{e^{\tan ^{-1} y} \tan ^{-1} y}{1+y^{2}} d y\) Putting \(\tan ^{-1} \mathrm{y}=\mathrm{z}\) in right hand side, \(\frac{1}{1+y^{2}} d y=d z\) \(\therefore \quad x^{\tan ^{-1} y}=\int e^{z} z d z\) \(= z^{z}-\int 1 \cdot e^{z} d z+C=\mathrm{ze}^{z}-e^{z}+c\) \(=\tan ^{-1} y \cdot e^{\tan ^{-1} y}-e^{\tan ^{-1} y}+C\) \(x^{\tan ^{-1} y}=\left(\tan ^{-1} y-1\right) e^{\tan ^{-1} y}+C\)
(B) : Given differential equation- \(\left(1+y^{2}\right) d x=\left(\tan ^{-1} y-x\right) d y\) Can be written as \(\frac{d y}{d x}=\frac{\tan ^{-1} y-x}{1+y^{2}}\) Or \(\quad \frac{d y}{d x}+\frac{1}{1+y^{2}} \cdot x=\frac{\tan ^{-1} y}{1+y^{2}}\) It is a linear differential of the form \(\frac{\mathrm{dx}}{\mathrm{dy}}+\mathrm{Px}=\mathrm{Q}\) Where \(\quad P=\frac{1}{1+y^{2}}, \quad Q=\frac{\tan ^{-1} y}{1+y^{2}}\) \(\therefore\) Integrating factor (IF) \(=\mathrm{e}^{\mathrm{Pdy}}\) \(=\mathrm{e}^{\int \frac{1}{1+\mathrm{y}^{2}} \mathrm{dy}}=\mathrm{e}^{\tan ^{-1} \mathrm{y}}\) \(\mathrm{x} . \mathrm{IF}=\int \mathrm{Q} \cdot \mathrm{IF}+\mathrm{C}\) So, solution is \(\mathrm{e}^{\tan ^{-1} y} \cdot x=\int \frac{e^{\tan ^{-1} y} \tan ^{-1} y}{1+y^{2}} d y\) Putting \(\tan ^{-1} \mathrm{y}=\mathrm{z}\) in right hand side, \(\frac{1}{1+y^{2}} d y=d z\) \(\therefore \quad x^{\tan ^{-1} y}=\int e^{z} z d z\) \(= z^{z}-\int 1 \cdot e^{z} d z+C=\mathrm{ze}^{z}-e^{z}+c\) \(=\tan ^{-1} y \cdot e^{\tan ^{-1} y}-e^{\tan ^{-1} y}+C\) \(x^{\tan ^{-1} y}=\left(\tan ^{-1} y-1\right) e^{\tan ^{-1} y}+C\)
(B) : Given differential equation- \(\left(1+y^{2}\right) d x=\left(\tan ^{-1} y-x\right) d y\) Can be written as \(\frac{d y}{d x}=\frac{\tan ^{-1} y-x}{1+y^{2}}\) Or \(\quad \frac{d y}{d x}+\frac{1}{1+y^{2}} \cdot x=\frac{\tan ^{-1} y}{1+y^{2}}\) It is a linear differential of the form \(\frac{\mathrm{dx}}{\mathrm{dy}}+\mathrm{Px}=\mathrm{Q}\) Where \(\quad P=\frac{1}{1+y^{2}}, \quad Q=\frac{\tan ^{-1} y}{1+y^{2}}\) \(\therefore\) Integrating factor (IF) \(=\mathrm{e}^{\mathrm{Pdy}}\) \(=\mathrm{e}^{\int \frac{1}{1+\mathrm{y}^{2}} \mathrm{dy}}=\mathrm{e}^{\tan ^{-1} \mathrm{y}}\) \(\mathrm{x} . \mathrm{IF}=\int \mathrm{Q} \cdot \mathrm{IF}+\mathrm{C}\) So, solution is \(\mathrm{e}^{\tan ^{-1} y} \cdot x=\int \frac{e^{\tan ^{-1} y} \tan ^{-1} y}{1+y^{2}} d y\) Putting \(\tan ^{-1} \mathrm{y}=\mathrm{z}\) in right hand side, \(\frac{1}{1+y^{2}} d y=d z\) \(\therefore \quad x^{\tan ^{-1} y}=\int e^{z} z d z\) \(= z^{z}-\int 1 \cdot e^{z} d z+C=\mathrm{ze}^{z}-e^{z}+c\) \(=\tan ^{-1} y \cdot e^{\tan ^{-1} y}-e^{\tan ^{-1} y}+C\) \(x^{\tan ^{-1} y}=\left(\tan ^{-1} y-1\right) e^{\tan ^{-1} y}+C\)
(B) : Given differential equation- \(\left(1+y^{2}\right) d x=\left(\tan ^{-1} y-x\right) d y\) Can be written as \(\frac{d y}{d x}=\frac{\tan ^{-1} y-x}{1+y^{2}}\) Or \(\quad \frac{d y}{d x}+\frac{1}{1+y^{2}} \cdot x=\frac{\tan ^{-1} y}{1+y^{2}}\) It is a linear differential of the form \(\frac{\mathrm{dx}}{\mathrm{dy}}+\mathrm{Px}=\mathrm{Q}\) Where \(\quad P=\frac{1}{1+y^{2}}, \quad Q=\frac{\tan ^{-1} y}{1+y^{2}}\) \(\therefore\) Integrating factor (IF) \(=\mathrm{e}^{\mathrm{Pdy}}\) \(=\mathrm{e}^{\int \frac{1}{1+\mathrm{y}^{2}} \mathrm{dy}}=\mathrm{e}^{\tan ^{-1} \mathrm{y}}\) \(\mathrm{x} . \mathrm{IF}=\int \mathrm{Q} \cdot \mathrm{IF}+\mathrm{C}\) So, solution is \(\mathrm{e}^{\tan ^{-1} y} \cdot x=\int \frac{e^{\tan ^{-1} y} \tan ^{-1} y}{1+y^{2}} d y\) Putting \(\tan ^{-1} \mathrm{y}=\mathrm{z}\) in right hand side, \(\frac{1}{1+y^{2}} d y=d z\) \(\therefore \quad x^{\tan ^{-1} y}=\int e^{z} z d z\) \(= z^{z}-\int 1 \cdot e^{z} d z+C=\mathrm{ze}^{z}-e^{z}+c\) \(=\tan ^{-1} y \cdot e^{\tan ^{-1} y}-e^{\tan ^{-1} y}+C\) \(x^{\tan ^{-1} y}=\left(\tan ^{-1} y-1\right) e^{\tan ^{-1} y}+C\)