Mark (√) against the correct answer in the following: The general solution of the DE $x^{2} \frac{d y}{d x}=x^{2}+x y+y^{2}$ is. A. $\tan ^{-1} \frac{\mathrm{y}}{\mathrm{x}}=\log \mathrm{x}+\mathrm{C}$ B. $\tan ^{-1} \frac{\mathrm{x}}{\mathrm{y}}=\log \mathrm{x}+\mathrm{C}$ C. $\tan ^{-1} \frac{\mathrm{y}}{\mathrm{x}}=\log \mathrm{y}+\mathrm{C}$ D. None of these

Home » Mark (√) against the correct answer in the following: The general solution of the DE is. A. B. C. D. None of these

Mark (√) against the correct answer in the following: The general solution of the DE $x^{2} \frac{d y}{d x}=x^{2}+x y+y^{2}$ is.
A. $\tan ^{-1} \frac{\mathrm{y}}{\mathrm{x}}=\log \mathrm{x}+\mathrm{C}$
B. $\tan ^{-1} \frac{\mathrm{x}}{\mathrm{y}}=\log \mathrm{x}+\mathrm{C}$
C. $\tan ^{-1} \frac{\mathrm{y}}{\mathrm{x}}=\log \mathrm{y}+\mathrm{C}$
D. None of these

CBSE | Class 12 | Linear Differential Equations | Maths | RS Aggarwal

Mark (√) against the correct answer in the following: The general solution of the DE $x^{2} \frac{d y}{d x}=x^{2}+x y+y^{2}$ is.
A. $\tan ^{-1} \frac{\mathrm{y}}{\mathrm{x}}=\log \mathrm{x}+\mathrm{C}$
B. $\tan ^{-1} \frac{\mathrm{x}}{\mathrm{y}}=\log \mathrm{x}+\mathrm{C}$
C. $\tan ^{-1} \frac{\mathrm{y}}{\mathrm{x}}=\log \mathrm{y}+\mathrm{C}$
D. None of these

Solution:

$\begin{array}{l} \mathrm{x}^{2} \frac{d y}{d x}=x^{2}+x y+y^{2} \\ \frac{d y}{d x}=1+\frac{y}{x}+\frac{y^{2}}{x^{2}} \end{array}$
Let $\mathrm{y}=\mathrm{vx}$
$\begin{array}{l} \mathrm{dy} / \mathrm{dx}=\mathrm{v}+\mathrm{xdv} / \mathrm{dx} \\ \begin{array}{l} 1+\mathrm{v}+\mathrm{v}^{2}=\mathrm{v}+\mathrm{x} \frac{d v}{d x} \\ 1+\mathrm{v}+\mathrm{v}^{2}=\mathrm{v}+\mathrm{x} \frac{d v}{d x} \\ 1+\mathrm{v}^{2}=\mathrm{x} \frac{d v}{d x} \\ \frac{d x}{x}=\frac{d v}{v^{2}+1} \end{array} \end{array}$
On integrating on both sides, we get
$\log x=\tan ^{-1} v+c$
$\tan ^{-1} \frac{y}{x}=\log x+c$ (required solution)