This answer is restricted. Please login to view the answer of this question.

Login Now**Solution:**

Given

\(\frac{dy}{dx} = \frac{x^2}{y^2}\) and given condition y'(0) = 2

\(y^2 \frac{dy}{dx} = x^2\) —— (i)

let v = y^{3}

Differentiating w.r.t x, we get

\(\frac{dv}{dx} = 3y^2 \frac{dy}{dx}\)

\(\frac{1}{3}\frac{dv}{dx} = y^2 \frac{dy}{dx}\)

So, Eq^{n} (i) becomes,

\(or, \enspace \frac{1}{3}\frac{dv}{dx} = x^2\)

\(or, \enspace \frac{dv}{dx} = 3x^2])

comparing it with \(\frac{dy}{dx} + Py = Q\), we get

P = 0 and Q = 3x^{2}

Here, The integral factor (I.F.) is

\(I.F. = e^{\int P dx} = 1\)

Now, the solution of the given differential equation is

y × I.F. = ∫ (Q × I.F.) dx + C

v = ∫ 3x^{2 }dx + c

y^{3} = x^{3} + c

Now,

Given condition is y(0) = 2 so,

2^{3} = 0^{3} + c

c = 8

Hence, the required solution is

y^{3} = x^{3} + 8

If you found any type of error on the answer then please mention on the comment or report an answer or submit your new answer.

Click here to submit your answer.

HAMROCSIT.COM

## Discussion