Private solution to a polynomial differential equation

[Karol]

The polynomial equation and it's private solution:
$$(1)~~ay''+by'+cy=f(x)=kx^n,~~y=A_0x^n+A_1x^{n-1}+...+A$$
If i, for example, take $f(x)=kx^3$ i get, after substituting into (1), an expression like $Ax^3+Bx^2+Cx+D$ , but that doesn't equal $kx^3$

 

[Simon Bridge]

Please show your working.

 

[pasmith]

The general solution of $$ay'' + by' + cy = ky$$ is $Ae^{r_1x} + Be^{r_2x}$ where $r_1 \neq r_2$ are the roots of $ar^2 + br + (c - k) = 0$. If the roots are equal then the general solution is instead $Ae^{r_1x} + Bxe^{r_1x}$.

You can therefore obtain $y(x) = x$ as a solution by taking $b = 0$ and $k = c$.

 

[Simon Bridge]

The general solution of $$ay'' + by' + cy = ky$$ is $Ae^{r_1x} + Be^{r_2x}$ where $r_1 \neq r_2$ are the roots of $ar^2 + br + (c - k) = 0$. If the roots are equal then the general solution is instead $Ae^{r_1x} + Bxe^{r_1x}$.

You can therefore obtain $y(x) = x$ as a solution by taking $b = 0$ and $k = c$.

But that is that the DE under consideration in post #1?

Compare:
$ay'' + by' + cy = ky$ with
$ay'' + by' + cy = kx^3$ ...

 

[Karol]

Please show your working.

$$y=A_0x^3+A_1x^2+A_2x+A,~~y'=3A_0x^2+2A_1x+A_2,~~y''=6A_0x+2A_1$$
$$ay''+by'+cy=6aA_0x+2aA_1+3bA_0x_2+6bA_1x+bA_2+cA_1x^2+cA_2x+cA=(cA_0)x^3+(3bA_0+cA_1)x^2+(6aA_0+6bA_1+cA_2)x+(2aA_1+bA_2+cA)$$
Is there a phrase about a polynomial with many members to equal only one with the highest power?

 

[Simon Bridge]

OK ... is that as far as you got?
You need an $=kx^3$ in there someplace?
Check the algebra on the others too ... then solve for the unknown coefficients.
(Probably easier to write the coefficients as A,B,C,D like you did in post #1)

 

[Karol]

You need an =$kx^3$ in there someplace?

$$(cA_0)x^3+(3bA_0+cA_1)x^2+(6aA_0+6bA_1+cA_2)x+(2aA_1+bA_2+cA)=kx^3$$

Check the algebra on the others too ... then solve for the unknown coefficients.

I
$$\left\{ \begin{array}{l} cA_0=k \\ 3bA_0+cA_1=0 \\ 6aA_0+6bA_1+cA_2=0 \\ 2aA_1+bA_2+cA=0 \end{array}\right.$$
$$\rightarrow~~A_0=\frac{k}{c},~~A_1=-\frac{3bk}{c^2},~~A_2=\left( \frac{3b_2c-a}{c^3} \right)6k,~~A=\frac{6abk}{c^3}-\left( \frac{3b_2c-a}{c^3} \right)6bk$$
Is that what you meant?

 

[Simon Bridge]

That's the idea... test for a=b=c=d=k=1 (say).

 

[Karol]

Thank you pasmith and Simon