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Here is what I have so far $y_g=C_1x^{1/2}+C_2x^{1/2}lnx$

We write

$y_p=U_1(x)x^{1/2}+U_2(x)x^{1/2}lnx$ $y_p'=x^{1/2}U_1'+x^{1/2}lnxU_2'+1/2x^{-1/2}U_1+(1/2x^{-1/2}lnx+x^{-1/2})U_2$

We choose to impose that $x^{1/2}U_1'+x^{1/2}lnxU_2'=0$

Then $y_p''=-1/4x^{-3/2}U_1+1/2x^{-1/2}U_1'+(1/2x^{-3/2}-1/4x^{-3/2}lnx)U_2+(1/2x^{-1/2}lnx+x^{-1/2})U_2'$

Subbing into the ODE and multiplying out we have


Now i'm not sure if my equations are correct but i'm pretty sure I've made a mistake because I think that somehow I show be able to take the final equation and the one that was imposed to 0 and take them away from each other but if they are right I have no idea how to do this.

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1 Answer 1

Why not derive the formulas $U_1 = \int \frac{-gy_2 dx}{W}$ and $U_2 = \int \frac{gy_1 dx}{W}$ where $W = y_1y_2'-y_2y_1'$ is the Wronskian of the fundamental solution set $\{ y_1,y_2 \}$ for the homogeneous ODE $4x^2y''+y=g=8\sqrt{x}$. Accepting that you did find the homogeneous solutions correctly we calculate: $$W=\sqrt{x}\biggl[\frac{1}{x} +\frac{1}{2\sqrt{x}}\ln(x)\biggr]-\frac{1}{2\sqrt{x}}\ln(x) = \frac{1}{\sqrt{x}} $$ where I have identified $y_1 = \sqrt{x}$ and $y_2 = \sqrt{x}\ln(x)$. It follows that: $$ U_1 = \int \sqrt{x} (-8\sqrt{x})(\sqrt{x}\ln(x))\, dx = -\int 8x^{3/2}\ln(x) \, dx$$ and $$ U_2 = \int \sqrt{x} (8\sqrt{x})(\sqrt{x})\, dx = \int 8x^{3/2} \, dx = \frac{16}{5}x^{5/2}.$$ To complete this you need to perform the integration for $U_1$ and write $y=c_1y_1+c_2y_2 + y_1U_1+y_2U_2$.

Some instructors prefer you complete the solution as you were. But, to do such is just to derive the formulas I state at the outset for different examples. My personal opinion is that it is easier to once derive the formulas and apply them. Of course with your method you do get experience solving systems of ODEs. This amounts to solving two equations for two unknowns ($U_1',U_2'$) then integrating. I can provide the derivation of the formulas if you wish.

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I think we have to do it the way that I am currently doing it. I've just got confused with my own working and not sure how to solve for the two unknowns from where i've got –  Adam Nov 10 '12 at 17:12
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