# Finding the general solution of a sixth degree differential equation

Find a differential equation whose solutions are $y_1 = e^{2x} + e^{-4x}\sin(3x)$ and $y_2 = e^{-2x} + 5e^{2x}$.

Am I supposed to assume that $y_1$ and $y_2$ can take the forms:

$y_1 = Ae^{2x} + e^{-4x}[C\cos(3x)+D\sin(3x)]$

$y_2 = Ee^{-2x} + Fe^{2x}$

I'm not sure where to go from here. Any advice?

Thank you!

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The title seems completely inappropriate. Would you like to edit it into something actually related to the question you ask? –  Gerry Myerson Oct 6 '12 at 10:16
@GerryMyerson: I agree. –  Mhenni Benghorbal Oct 6 '12 at 17:33

You can use the annihilator method. For example, the function ${\rm e}^{\alpha x}$ has the annihilator $D-\alpha$, where $D=\frac{d}{dx}$. That means

$$(D-\alpha){\rm e}^{\alpha x}= \alpha {\rm e}^{\alpha x}-\alpha {\rm e}^{\alpha x} = 0 \,.$$

Now, since $y_1$ and $y_2$ are solutions, then $y_1+y_2$ is a solution too.

$$y(x) = 6e^{2x} + e^{-4x} \sin(3x) + e^{-2x}= 6e^{2x}+ \frac{1}{2i}e^{(-4+3i)x}- \frac{1}{2i}e^{(-4-3i)x} + e^{-2x}\,.$$

To annihilate the above equation, we apply the above annihilators to both sides of the equation

$$(D+2)(D-(-4-3i))(D-(-4+3i))(D-2)y(x) = 0 \,.$$

Multiplying and simplifying the left hand side gives a differential equation of fourth order

$$({D}^{4}+8\,{D}^{3}+21\,{D}^{2}-32\,D-100)y(x)=0$$

$$\Rightarrow y^{(4)}+8y^{(3)}+21y^{(2)}-32y'-100y=0\,.$$

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this is so cool! –  Ant Feb 28 '14 at 9:00

$$y_1=e^{2x}+\frac12\left(e^{(-4+3i)x}-e^{(-4-3i)x}\right)$$

Now take a differential equation of the form

$$ay+by'+cy''+y'''=0$$

Sub in $y=e^{rx}$:

$$e^{rx}(a+br+cr^2+r^3)=0$$

So we want to find a cubic $r^3+cr^2+br+a$ that has its roots as the coefficients in the above exponents. So we chose

$$(r-2)(r+4-3i)(r+4+3i)=0$$

which will generate each exponential in the expression for $y_1$ as solutions. Expanding, we get

$$r^3+6r^2+9r-50=0$$

So that $y_1$ is a solution of

$$-50y+9y'+6y''+y'''=0$$

The case for $y_2$ proceeds similarly, with a 2nd order differential equation instead.

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I chose a linear differential equation of third order because linear differential equations generate combinations of exponentials, sines and cosines (really just exponential functions to complex powers) as solutions, and because we had three terms after expanding in terms of exponentials. –  Robert Mastragostino Oct 6 '12 at 6:16
Where does the 1/2 come from in the first $y_1$? Also, why do we substitute $e^{rx}$ into the differential equation? –  Bob John Oct 6 '12 at 6:19
It's an identity from complex analysis: $\sin(x)=\frac12(e^{ix}-e^{-ix})$. And the idea is that the derivative of an exponential is still an exponential: when we solve linear differential equations, we use this substitution to turn our problem into solving an algebraic equation (in $r$) instead, which we can do with known methods. –  Robert Mastragostino Oct 6 '12 at 6:24
Ok, that makes sense. But why is $i = -4\pm 3i$? More specifically, why are we assigning $i$ the value of the complex root, is that what it represents? –  Bob John Oct 6 '12 at 6:53