Question

What is a general solution to the differential equation `y'+2xy=2x^3`?

Answer 1

The Gen. Soln. is,

`ye^(x^2)=(x^2-1)e^(x^2)+C`, or, `y=x^2-1+Ce^(-x^2)`.

Explanation:

This is a Linear Diff. Eqn. `: dy/dx+P(x)*y=Q(x)...............(1)`,

where, `P(x)` and `Q(x)` are funs. of variable `x`.

To find its Gen. Soln. , we have to multiply it by, Integrating

Factor given by, `I.F.=e^(intP(x)dx)`.

Comparing the given diff. eqn. with `(1)`, we have,

`P(x)=2xrArrintP(x)dx=int2xdx=x^2rArr I.F.=e^(x^2)`.

Multiplying the eqn. throughout by `I,F.`, we get,

`e^(x^2)dy/dx+2xye^(x^2)=2x^3e^(x^2)`

`:. e^(x^2)d/dx(y)+yd/dx(e^(x^2))=2x^3e^(x^2)`.

`:. d/dx(y*e^(x^2))=2x^3e^(x^2)`

`:. y*e^(x^2)=int2x^3e^(x^2)dx............(2)`

To evaluate the integral, say `I`, on `R.H.S.`, we will subst.,

`x^2=t`, so that, `2xdx=dt`.

`:.I=intte^tdt`

`=tinte^tdt-int[(d/dt(t)*inte^tdt]dt............[IBP]`

`=te^t-inte^tdt=te^t-e^t=(t-1)e^t=(x^2-1)e^(x^2)+C`.

By `(2)`, then, the Gen. Soln. is,

`ye^(x^2)=(x^2-1)e^(x^2)+C`, or, `y=x^2-1+Ce^(-x^2)`.

Answer 2

`y = x^2 - 1 + C e^(-x^2)`

Explanation:

`y'+2xy=2x^3`

this is not separable. we can use an integrating factor `eta(x)`

here `eta = exp (int 2x \ dx) = e^(x^2)`

so
`e^(x^2) y'+e^(x^2) 2xy=e^(x^2) 2x^3`

or

`d/dx (e^(x^2) y)=e^(x^2) 2x^3`

`e^(x^2) y = int \ e^(x^2) 2x^3 dx`

`e^(x^2) y = int \ d/dx ( e^(x^2) ) x^2 dx`

using IBP

`e^(x^2) y = x^2 e^(x^2) - int \ e^(x^2) d/dx( x^2) dx`

`e^(x^2) y = x^2 e^(x^2) - 2 int \ x e^(x^2) dx`

and spotting the pattern in the integrand

`e^(x^2) y = x^2 e^(x^2) - 2 int \ d/dx(1/2 e^(x^2)) dx`

`e^(x^2) y = x^2 e^(x^2) - e^(x^2) + C`

`y = x^2 - 1 + C e^(-x^2)`