The solution of the differential equation x²(y+1) dy/dx + y²(x+1)² = 0, when y(1)=2, is [AP EAPCET 24 May 2025 S1]

Correct Answer is : log⁡∣12x2y∣=1x+1y−x−12≤ft|1/2 x² y| = 1/x + 1/y - x - 1/2log21x2y=x1+y1−x−21

≤ft|1/4 x² y| = 1/x + 2/y + x - 1

≤ft|1/2 x² y| = 1/x + 1/y - x - 1/2

≤ft|1/3 x² y| = 1/x + 1/y - x + 1/2

Correct Answer is : log⁡∣12x2y∣=1x+1y−x−12≤ft|1/2 x² y| = 1/x + 1/y - x - 1/2log21x2y=x1+y1−x−21

Test Index

Differential Equations

Section: Mathematics

Question : 7 of 48

Marks: +1, -0

x^2(y+1) \frac{dy}{dx} + y^2(x+1)^2 = 0

y(1)=2

$\log|x^2 y| = \frac{2}{x} + \frac{1}{y} + x - 1$
$\log\left|\frac{1}{4} x^2 y\right| = \frac{1}{x} + \frac{2}{y} + x - 1$
$\log\left|\frac{1}{2} x^2 y\right| = \frac{1}{x} + \frac{1}{y} - x - \frac{1}{2}$
$\log\left|\frac{1}{3} x^2 y\right| = \frac{1}{x} + \frac{1}{y} - x + \frac{1}{2}$

Solution:

Given, differential equation is

x^2(y+1) \frac{dy}{dx} + y^2(x+1)^2 = 0, y(1)=2

\Rightarrow x^2(y+1) \frac{dy}{dx} = -y^2(x+1)^2

\Rightarrow \frac{y+1}{y^2} dy = -\frac{(x+1)^2}{x^2} dx

Integrating to both sides, we get

\int \frac{y+1}{y^2} dy = -\int \frac{(x+1)^2}{x^2} dx

\Rightarrow \int \left( \frac{1}{y} + \frac{1}{y^2} \right) dy

= -\int \left( 1 + \frac{2}{x} + \frac{1}{x^2} \right) dx

\Rightarrow \ln|y| - \frac{1}{y} = -\left( x + 2\ln|x| - \frac{1}{x} \right) + C,

where

C=

constant

\Rightarrow \ln|y| + 2\ln|x| = \frac{1}{y} + \frac{1}{x} - x + C

\Rightarrow \ln|x^2 y| = \frac{1}{y} + \frac{1}{x} - x + C \cdot\cdot\cdot\cdot\cdot\cdot\cdot \text{(i)}

Given,

y(1)=2

, so we get

\Rightarrow \ln|1^2 \cdot 2| = \frac{1}{2} + \frac{1}{1} - 1 + C

\Rightarrow \ln|2| = \frac{1}{2} + C \Rightarrow c = \ln|2| - \frac{1}{2}

From, Eq. (i), we get

\ln|x^2 y| = \frac{1}{y} + \frac{1}{x} - x + \ln|2| - \frac{1}{2}

\Rightarrow \ln|x^2 y| - \ln|2| = \frac{1}{y} + \frac{1}{x} - x - \frac{1}{2}

\Rightarrow \ln\left|\frac{1}{2} x^2 y\right| = \frac{1}{y} + \frac{1}{x} - x - \frac{1}{2}

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