Question

Solve the differential equation:

`(x + 5y^2) dy/dx = y` when x = 2 and y = 1

Answer 1

Given, differential equation is:

`(x + 5y^2) dy/dx = y`

`dy/dx = y/(x + 5y^2)`

⇒ `dx/dy = (x + 5y^2)/y`

⇒ `dx/dy - x/y = 5y`

Compare with `dx/dy + Px = Q`

Here, `P = -1/y, Q = 5y`

I.F. = `e^(int P dy)`

= `e^(int - 1/y  dy)`

= `e^(-log y)`

I.F. = `1/y`

Solution is

x × I.F. = ∫ Q × I.F. dy

`x/y = int 5y xx 1/y dy`

⇒ `x/y = 5y + C`

When x = 2 and y = 1

`2/1 = 5 xx 1 + C`

⇒ C = – 3

Now, `x/y = 5y - 3`

So, x = 5y² – 3y.