The solution of the equation x² y-x³ d y/d x=y^{4} x, where y(0)=1, is

Correct Answer is : x3=3y3sin⁡xx³=3 y³ xx3=3y3sinx

Correct Answer is : x3=3y3sin⁡xx³=3 y³ xx3=3y3sinx

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MHTCET 25 Apr 2025 Shift 1 Paper

Question : 123 of 150

Marks: +1, -0

x^{2} y-x^{3} \frac{d y}{d x}=y^{4} \cos x

y(0)=1

, is

Solution:

Concept:

This is a first-order differential equation reducible to linear form using the substitution

v = \frac{1}{y^3}

Explanation:

Given equation:

x^{2}y - x^{3}\frac{dy}{dx} = y^{4}\cos x

Rearrange:

x^{3}\frac{dy}{dx} - x^{2}y = -y^{4}\cos x

Divide by

x^{3}y^{4}

\frac{-1}{y^{4}}\frac{dy}{dx} + \frac{1}{xy^{3}} = \frac{1}{x^{3}}\cos x

Let

v = \frac{1}{y^{3}}

Then:

\frac{-1}{y^{4}}\frac{dy}{dx} = \frac{1}{3}\frac{dv}{dx}

Substitute into the equation:

\frac{1}{3}\frac{dv}{dx} + \frac{v}{x} = \frac{1}{x^{3}}\cos x

Multiply by

3

\frac{dv}{dx} + \frac{3v}{x} = \frac{3}{x^{3}}\cos x

This is linear in

v

. The integrating factor is:

\text{I.F.} = e^{\int \frac{3}{x}\,dx} = e^{3\ln x} = x^{3}

So the solution is:

v \cdot x^{3} = \int \frac{3}{x^{3}}\cos x \cdot x^{3}\,dx + c

\Rightarrow \frac{x^{3}}{y^{3}} = 3\int \cos x\,dx + c

\Rightarrow \frac{x^{3}}{y^{3}} = 3\sin x + c

Given

y(0) = 1

, i.e.

x = 0

y = 1

\frac{0}{1} = 3\sin 0 + c \Rightarrow c = 0

Hence:

\frac{x^{3}}{y^{3}} = 3\sin x

\Rightarrow x^{3} = 3y^{3}\sin x

Answer:

Option B:

x^{3} = 3y^{3}\sin x

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