What is the solution of the differential equation (d x)/(d y)+(x)/(y)-y^2=0 ? where c is an arbitrary constant.

Correct answer: C. 4 x y=y^4+c aligned & (d x)/(d y)+(x)/(y)-y^2=0 & (d x)/(d y)+(x)/(y)=y^2 aligned This is a linear differential equation of the form dx dy +P_1 x=Q _1 Here, P= 1 y and Q=y^2 I.F. = ^ ∫ P d y = ^ ∫ (1)/(y) d y = ^ y =y So, required solution is aligned & x · y=∫ y^2 · y d y+c & x y=∫ y^3 d y+c & x y=(y^4)/(4)+c & 4 x y=y^4+c aligned Option (C) is correct.

NDA 2015

Exam: NDA
Year: 2015
Subject: Mathematics
Chapter: Differential Equations

NDA 2015 · Mathematics · Differential Equations

What is the solution of the differential equation (d x)/(d...

What is the solution of the differential equation

\frac{d x}{d y}+\frac{x}{y}-y^2=0

? where

c

is an arbitrary constant.

Options

A $x y=x^4+c$
B $x y=y^4+c$
C $4 x y=y^4+c$
D $3 x y=y^3+c$

Correct answer

C. $4 x y=y^4+c$

Step-by-step solution

\begin{aligned} & \frac{d x}{d y}+\frac{x}{y}-y^2=0 \\ & \frac{d x}{d y}+\frac{x}{y}=y^2 \end{aligned}

This is a linear differential equation of the form

\frac{\mathrm{dx}}{\mathrm{dy}}+\mathrm{P}_1

\mathrm{x}=\mathrm{Q}

_1

Here,

\mathrm{P}=\frac{1}{\mathrm{y}}

and

\mathrm{Q}=\mathrm{y}^2\therefore \quad \text{ I.F. }=\mathrme^{\int P d y}=\mathrme^{\int \frac{1}{y} d y}=\mathrme^{\log y}=y

So, required solution is

\begin{aligned} & x \cdot y=\int y^2 \cdot y d y+c \\ & x y=\int y^3 d y+c \\ & x y=\frac{y^4}{4}+c \\ & 4 x y=y^4+c \end{aligned}

\therefore \quad$ Option (C) is correct.

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