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# Solution for The Solution of the Differential Equation D Y D X = X 2 + X Y + Y 2 X 2 , is (A) Tan − 1 ( X Y ) = Log Y + C (B) Tan − 1 ( Y X ) = Log X + C (C) Tan − 1 ( X Y ) = Log X + C - CBSE (Science) Class 12 - Mathematics

ConceptGeneral and Particular Solutions of a Differential Equation

#### Question

The solution of the differential equation $\frac{dy}{dx} = \frac{x^2 + xy + y^2}{x^2}$, is
(a) $\tan^{- 1} \left( \frac{x}{y} \right) = \log y + C$
(b) $\tan^{- 1} \left( \frac{y}{x} \right) = \log x + C$
(c) $\tan^{- 1} \left( \frac{x}{y} \right) = \log x + C$
(d) $\tan^{- 1} \left( \frac{y}{x} \right) = \log y + C$

#### Solution

$\left( b \right) \tan^{- 1} \left( \frac{y}{x} \right) = \log x + C$
We have,
$\frac{dy}{dx} = \frac{x^2 + xy + y^2}{x^2}$  ................(1)
This is homogenous differential equation.
$\text{ Let }y = vx$
$\Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}$
$\text{ Now, putting }\frac{dy}{dx} = v + x\frac{dv}{dx}\text{ and }y = vx\text{ in }\left( 1 \right),\text{ we get }$
$v + x\frac{dv}{dx} = \frac{x^2 + x^2 v + x^2 v^2}{x^2}$
$\Rightarrow v + x\frac{dv}{dx} = 1 + v + v^2$
$\Rightarrow x\frac{dv}{dx} = 1 + v^2$
$\Rightarrow \left( \frac{1}{1 + v^2} \right)dv = \frac{1}{x}dx$
Integrating both sides we get,
$\int\frac{1}{1 + v^2}dv = \int\frac{1}{x}dx$
$\Rightarrow \tan^{- 1} v = \log x + C$
$\Rightarrow \tan^{- 1} \left( \frac{y}{x} \right) = \log x + C$
Is there an error in this question or solution?

#### Video TutorialsVIEW ALL [2]

Solution The Solution of the Differential Equation D Y D X = X 2 + X Y + Y 2 X 2 , is (A) Tan − 1 ( X Y ) = Log Y + C (B) Tan − 1 ( Y X ) = Log X + C (C) Tan − 1 ( X Y ) = Log X + C Concept: General and Particular Solutions of a Differential Equation.
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