# Solve y’ = x2/y2, y(0) = 2

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