Solve y’ = x2/y2, y(0) = 2

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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 = y3

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, Eqn (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 = 3x2

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 = ∫ 3xdx + c

y3 = x3 + c

Now,

Given condition is y(0) = 2 so,

23 = 03 + c

c = 8

Hence, the required solution is

y3 = x3 + 8

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