Particular Solution of the ODE $y''-6y'+10y=e^{3x}$

Question

I have to find the particular solution of the ODE $y''-6y'+10y=e^{3x}$ using, what our professor calls The Integration Formula which is:

For the non-homogeneous ODE: $$p_0(x)y''+p_1(x)y'+p_2(x)y=r(x)$$

The Integration Formula is: $$\int_{}^{x} H(x,t)\frac{r(t)}{p_0(t)} \,dt,$$

Where $$H(x,t)=\frac{\begin{vmatrix} y_1(t) & y_2(t) \\ y_1(x) & y_2(x) \end{vmatrix}}{\begin{vmatrix} y_1(t) & y_2(t) \\ y_1'(t) & y_2'(t) \end{vmatrix}}.$$

where $y_1$ and $y_2$ are the solutions of $y''-6y'+10y=0$, namely $y_1(x)=\cos x$ and $y_2(x)=\sin x$.

My Approach:

$$\int_{}^{x} H(x,t)\frac{r(t)}{p_0(t)} \,dt=\int_{}^{x}e^{3t}(\cos t \sin x-\sin t \cos x)\,dt$$

I Break this Integral into two: $$\int_{}^{x}e^{3t}\cos t \sin x\,dt +\int_{}^{x}-e^{3t}\sin t \cos x \, dt$$

Integrating both integrals by parts and then replacing with $x$ wherever I have $t$, gives: $$\frac{e^{3x}(\sin x+3\cos x) \sin x}{10}-\frac{e^{3x}(3\sin x-\cos x)\cos x}{10}=\frac{e^{3x}}{10}$$

Thus I find that the Particular Solution is $\frac{e^{3x}}{10}$. However, I know for a fact that the particular solution is $e^{3x}$ (verified by calculators as well). I am pretty sure that my integral calculations are correct, however I have my doubts for the original formula I had to use, namely, $$\int_{}^{x} H(x,t)\frac{r(t)}{p_0(t)} \,dt.$$

Can someone point out my mistake?

Edit:

My solutions for $y''-6y'+10y=0$ were wrong and in fact they were $y_1(x)=e^{3x}\cos x$ and $y_2(x)=e^{3x}\sin x$. Thus,

$$H(x,t)=\frac{e^{3x}(\cos t \sin x - \sin t \cos x)}{e^{3t}}$$.

The Integral is then: $$\int_{}^{x} H(x,t)\frac{r(t)}{p_0(t)} \,dt=\int_{}^{x}e^{3x}(\cos t \sin x-\sin t \cos x)\,dt=e^{3x}(\sin^2 x+\cos^2 x)=e^{3x}$$.

Thus indeed the particular solution is $$y_0=e^{3x}$$.

Answer 1

The formula you wrote is a correct way of writing down what is usually called the method of variation of parameters. When you write it like that, the name seems rather mysterious. The idea behind the name is that the solution is of the form $y_p(x)=c_1(x) y_1(x) + c_2(x) y_2(x)$ (so you "take the homogeneous solution but let the parameters vary"). The method chooses these functions $c_1,c_2$.

Anyway, the actual problem with your solution is simply that you solved the homogeneous equation incorrectly: $i^2-6i+10 \neq 0$.

Answer 2

When the RHS of the equation contains an exponential, the best way is to begin by eliminating it by putting

$$y=ze^{3x}$$ and $$y'=(z'+3z)e^{3x}$$ $$y''=(z''+6z'+9z)e^{3x}$$

the equation becomes

$$z''+6z'+9z-6z'-18z+10z=1$$ or $$z''+z=1$$ Now you can apply your method to this easier equation.

from here we can get the particular solution $$z_p=1\;\text{ and } \;y_p=e^{3x}$$

Answer 3

The method your professor advocates is known (at least to me) as Variation of Parameters. I believe in posting complete answers to queries here rather than hints or tips so I wil walk you through exactly what you need to do. You correctly pointed out that you started by solving the homogeneous differential equation and using those (presumably linearly independent) functions to create the non-homogeneous solution.

Solving the homogeneous part, we find that \begin{align*} \lambda^2-6\lambda+10=0 \implies \lambda_1 = i + 3 \lor \lambda_2 = -i+3 \end{align*} with multiplicities of $1$ each. Thus the partial solution becomes: \begin{align*} y = C_1 e^{3x} \sin x+ C_2 e^{3x} \cos x \longleftrightarrow \{ e^{3x}\sin x, e^{3x} \cos x\} \end{align*} the pair of the functions in question. It is trivial to show that these are linearly independent as well.

The Wronskian of the two is \begin{align} \begin{vmatrix} e^{3x}\sin x & e^{3x} \cos x\\ e^{3x} (3 \sin x + \cos x) & -e^{3x} (\sin x -3 \cos x) \end{vmatrix} = \ldots = -e^{6x} \neq 0 \end{align} which was to be expected from linear independence. So from your method the solution to the nonhomogeneous part (partial solution) is given by the integral \begin{align} y_{partial}\left(x \right) = - {y_1}\int{{\frac{{{y_2}g\left( x \right)}}{{W\left( {{y_1},{y_2}} \right)}}\,dx}} + {y_2}\int{{\frac{{{y_1}g\left( x \right)}}{{W\left( {{y_1},{y_2}} \right)}}\,dx}}. \end{align}

Putting in the necessary parts we get \begin{align} y_{partial}\left(x \right) &= -e^{3x}\sin x \int \dfrac{e^{3x} \cos x \cdot e^{3x}}{-e^{6x}} dx + e^{3x} \cos x \int \dfrac{e^{3x} \sin x \cdot e^{3x}}{-e^{6x}}dx\\ &= e^{3x} \sin x \int \cos x dx - e^{3x} \cos x \int \sin x dx\\ &= e^{3x} \sin^{2} x + e^{3x} \cos ^{2} x\\ &= e^{3x} \end{align} as we wanted.