Find the solution to the initial Value problem: X^(2)y'+3xy=x^(1/2), y(1)=2

Can you please explain it step by step, I was blessed with a professor who can't teach

thank you in advance

Okay lets do it:

Lets assume the function y = uv, where u = u(x) and v = v(x), because of that:

y' = u'v + v'u

By knowing these lets first divide the main equation by x^2:

y' + 3x/y = x^(-3/2)

Now lets substitute our assumptions to this equation:

u'v + v'u + 3uv/x = x^(-3/2)

u( v' + 3v/x ) + u'v = x^(-3/2)

We can put this equation into two parts now to make system of equations:

v' + 3v/x = 0 [1]

u'v = x^(-3/2) [2]

Lets solve the first equation, by assuming that dv/dx = v'

dv/dx = - 3v/x

By seperating the variables we get:

dv/v = -3dx/x => lnv = -3lnx + C, we chose this constant to be zero, we will get another constant pretty soon.

By solving equation above we get: v = x^(-3)

Ok lets get to the equation [2] and substitute v to it:

u'x^(-3) = x^(-3/2)

u' = x^(3/2) => du/dx = x^(3/2)

And by separating variables we get: du = x^(3/2) dx And by integrating receive:

u = 2/5* x^(5/2) + C

Now remember we chose y = uv, now we got u and v, lets substitute this here to obtain the final equation:

y = u*v = (2/5* x^(5/2) + C)*x^-3 = 2/5* x^(-1/2) + Cx^(-3)

Now to determine the coefficient C for the case of y(1)=2, we substitute x=1 and y=2 to the equation:

2 = 2/5 + C => C = 2-2/5 = 8/5 and the equation becomes:

y = 2/5* x^(-1/2) + 8/5* x^(-3)

oo
Infinite