Is this the general solution?

[mathmari]

Hey!

I want to solve the differential equation $y'''(x)-2y''(x)+y'(x)=e^x$.

I have done the following:

Consider the homogeneous equation $y'''(x)-2y''(x)+y'(x)$.

$k^3-2k^2+k=0 \Rightarrow k=0 \text{ single root } , k=1 \text{ double root } $

So, the solution of the homogeneous problem is $y_h(x)=c_1+c_2e^x+c_3xe^x$. Since $1$ is a root of the characteristic polynomial of multiplicity $2$, we consider that the partial solution is of the form $y_p=(Ax^2+Bx+C)e^{x}$.

Finding the derivatives $y'_p,y''_p,y'''_p$ and replacing it at the problem I get $A=\frac{1}{2}$.

Is this correct? (Wondering) Is then the general solution the following? (Wondering)

$$y(x)=y_h(x)+y_p(x) \\ =c_1+c_2e^x+c_3xe^x+\left (\frac{1}{2}x^2+Bx+C\right )e^{x} \\ =c_1+(c_2+C)e^x+(c_3+B)xe^x+\frac{1}{2}x^2e^x$$

 

[MarkFL]

When you use the method of undetermined coefficients to obtain the particular solution, you should find numerical values for all of the parameters $A,\,B,\,C$.

 

[Opalg]

I want to solve the differential equation $y'''(x)-2y''(x)+y'(x)=e^x$.

...

Is then the general solution the following? (Wondering)

$$y(x)=y_h(x)+y_p(x) \\ =c_1+c_2e^x+c_3xe^x+\left (\frac{1}{2}x^2+Bx+C\right )e^{x} \\ =c_1+(c_2+C)e^x+(c_3+B)xe^x+\frac{1}{2}x^2e^x$$

Yes it is! But it would be better to check the answer yourself rather than ask MHB to do it for you.

In fact, writing $L,M,N$ for the constants $c_1$, $c_2+C$ and $c_3+B$, your proposed solution is $$y = L + Me^x + Nxe^x + \tfrac12x^2e^x.$$ You can then differentiate this three times to get $$y' = Me^x + N(1+x)e^x + (\tfrac12x^2+x)e^x,$$ $$y'' = (M+N)e^x + N(1+x)e^x + (\tfrac12x^2+2x+1)e^x,$$ $$y''' = (M+2N)e^x + N(1+x)e^x + (\tfrac12x^2+3x+3)e^x.$$ It is then a simple matter to check that $y''' - 2y'' + y' = e^x.$ So your solution works, and since it contains three arbitrary constants it must be the general solution to this third-order equation.