Question

How do you solve `y'=-xy+sqrty` given `y(0)=0`?

Answer 1

`y=y(0)=1/4pie^(-1/2x^2)"erfi"^2(1/2x)`

Explanation:

We have `y'=-xy+sqrty` or `dy/dx=y^(1/2)-xy`.

Let `v=y^(1/2)` and `(dv)/dx=1/2y^(-1/2)dy/dx`

Now

`y^(-1/2)dy/dx=1-xy^(1/2)=1-vx`

so

`(dv)/dx=1/2(1-vx)=1/2-1/2vx`

`rArr(dv)/dx+1/2vx=1/2`

This is now a linear first-order ordinary differential equation of the form

`(dv)/(dx)+vP(x)=Q(x)`

We solve this using an integrating factor:

`mu=e^(int1/2xdx)=e^(1/4x^2)`

So the general solution is

`ve^(1/4x^2)=int1/2e^(1/4x^2)dx=1/2(sqrtpi "erfi"(1/2x)+"c")`

Solving for `v`

`v=1/2e^(-1/4x^2)(sqrtpi"erfi"(1/2x)+"c")`

But `v=sqrty` so

`sqrty=1/2e^(-1/4x^2)(sqrtpi"erfi"(1/2x)+"c")`

and

`y=1/4e^(-1/2x^2)(sqrtpi"erfi"(1/2x)+"c")^2`

Now, we have the condition `y(0)=0`

So

`y(0)=1/4e^0(sqrtpi"erfi"(0/2)+"c")^2=0`

`rArr1/4c^2=0rArrc=0`

So the final solution is

`y=y(0)=1/4pie^(-1/2x^2)"erfi"^2(1/2x)`