Question

x²y''-2xy'+2y=4x3
find the general solution of this differential equation.Last of term is 4x(cube).

Answer 1

what ways can we use

Answer 2

cauchy-euler method

Answer 3

x power to alpha

Answer 4

so we can to assume the solution is
\[y=x^r \\ \text{ and so the following follows } \\ y'=r x^{r-1} \\ y''=r(r-1)x^{r-2} \\ \text{ plug \in } \\ x^2 \cdot r (r-1)x^{r-2}-2x \cdot r x^{r-1}+2 \cdot x^{r}=4x^3\]

Answer 5

oops that 4x^3 was suppose to be 0

Answer 6

\[y=x^r \\ \text{ and so the following follows } \\ y'=r x^{r-1} \\ y''=r(r-1)x^{r-2} \\ \text{ plug \in } \\ x^2 \cdot r (r-1)x^{r-2}-2x \cdot r x^{r-1}+2 \cdot x^{r}=0\]
solve for r

Answer 7

ok i solved it with cauchy euler method

Answer 8

so you are saying you already got the homoegenous solution'?

Answer 9

yes and i also found partial solution

Answer 10

can we use reduction of order after the euler thing?

Answer 11

i do not know but with cauchy it will be found the solution

Answer 12

\[x^2y''-2xy'+2y=4x^3\]
Let \(x=e^t\), or \(t=\ln x\), so that \(y(x)=y(e^t)=z(t)\). This means
\[y'=\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{x}\frac{\mathrm{d}z}{\mathrm{d}t}=\frac{z'}{x}\\
y''=\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{1}{x^2}\left(\frac{\mathrm{d}^2z}{\mathrm{d}t^2}-\frac{\mathrm{d}z}{\mathrm{d}t}\right)=\frac{z''-z'}{x^2}\]
Substituting into the original ODE, you have
\[(z''-z')-2z'+2z=4e^{3t}\implies z''-3z'+2z=4e^{3t}\]
You seem to already have the homogeneous solution worked out, but I'll include it here just for the sake of completeness. The characteristic solution and its roots are
\[r^2-3r+2=(r-2)(r-1)=0\implies r_1=1\quad\text{and}\quad r_2=2\]so you know that two solutions are \(C_1e^t\) and \(C_2e^{2t}\).

For the nonhomogeneous part, you can try a particular solution of the form \(z_p=Ae^{3t}\), which has derivatives
\[{z_p}'=3Ae^{3t}\quad\text{and}\quad{z_p}''=9Ae^{3t}\]
Substituting into the ODE, you get
\[9Ae^{3t}-3\left(3Ae^{3t}\right)+2Ae^{3t}=4e^{3t}\]which you can then solve for \(A\) and find \(z_p\).

Keep in mind that the original ODE is in \(y\), not \(z\), so getting the final solution is just a matter of replacing \(t\) with \(\ln x\) (and simplifying, if you so choose).