# Thread: Differential Equation

1. ## Differential Equation

Could someone just point me in the right direction for a general solution to the following question?

$\displaystyle \frac{d^2y}{dx^2} - 2\frac{d^2y}{dx^2} + 10y = x^2e^{2x}$

Many thanks.

2. Originally Posted by pikminman
Could someone just point me in the right direction for a general solution to the following question?

$\displaystyle \frac{d^2y}{dx^2} - 2\frac{d^2y}{dx^2} + 10y = x^2e^{2x}$

Many thanks.
The general solution of this is of the form $\displaystyle y(x)=g(x)+h(x)$, where $\displaystyle g(x)$ is the general solution of the homogeneous equation:

$\displaystyle \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 10y = 0$

and $\displaystyle h(x)$ is a particular solution of the original equation. I would try a solution of the form $\displaystyle h(x)=P(x)e^{2x}$ where $\displaystyle P(x)$ is a quadratic in $\displaystyle x$

CB

3. Originally Posted by CaptainBlack
The general solution of this is of the form $\displaystyle y(x)=g(x)+h(x)$, where $\displaystyle g(x)$ is the general solution of the homogeneous equation:

$\displaystyle \frac{d^2y}{dx^2} - 2\frac{d^2y}{dx^2} + 10y = 0$

and $\displaystyle h(x)$ is a particular solution of the original equation. I would try a solution of the form $\displaystyle h(x)=P(x)e^{2x}$ where $\displaystyle P(x)$ is a quadratic in $\displaystyle x$

CB
I am not that good at maths. Could you explain in a slightly more babyish fashion please?

4. Originally Posted by pikminman
I am not that good at maths. Could you explain in a slightly more babyish fashion please?
I suggest you read this tutorial - http://www.mathhelpforum.com/math-he...-tutorial.html

In particular, the part about "Method of Undetermined Coefficients".

Do you know how to solve for the general solution for this?

$\displaystyle \frac{d^2y}{dx^2} - 2\frac{d^2y}{dx^2} + 10y = 0$

5. Originally Posted by pikminman
I am not that good at maths. Could you explain in a slightly more babyish fashion please?
Your aux equation would be

$\displaystyle r^2 -2r^2 + 10 = 0$

$\displaystyle -r^2 = -10$

$\displaystyle r^2 = 10$

$\displaystyle r = ^+_- \sqrt{10}$

giving you the solution of

$\displaystyle c_{1}e^{-\sqrt{10}x} + c_{2}e^{\sqrt{10}x}$

do you know how to do the next step?

6. Originally Posted by 11rdc11
Your aux equation would be

$\displaystyle r^2 -2r^2 + 10 = 0$

$\displaystyle -r^2 = -10$

$\displaystyle r^2 = 10$

$\displaystyle r = ^+_- \sqrt{10}$

giving you the solution of

$\displaystyle c_{1}e^{-\sqrt{10}t} + c_{2}e^{\sqrt{10}t}$

do you know how to do the next step?
Not really...

Oh! Yes, I meant to say $\displaystyle \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 10y = 0$ not $\displaystyle \frac{d^2y}{dx^2} - 2\frac{d^2y}{dx^2} + 10y = 0$ Sorry...

7. $\displaystyle y_{p} = (ax^2+bx + c)e^{2x}$

$\displaystyle y'_{p} =(2ax+b)(e^{2x}) +2e^{2x}(ax^2+bx+c)= e^{2x}(2ax+b +2(ax^2 +bx+c))$

$\displaystyle = e^{2x}(2ax + b + 2ax^2 +2bx +2c) = e^{2x}(2ax^2 +(2a+2b)x +b +2c)$

$\displaystyle y''_{p} = 2e^{2x}(2ax^2 +(2a+2b)x +b +2c) +e^{2x}(4ax +2a +2b)$

$\displaystyle = e^{2x}(4ax^2 +(4a+4b)x +2b +4c +4ax +2a +2b) = e^{2x}(4ax^2 +(8a+4b)x +4b +4c +2a)$

plugging it all back into the original equation gives

$\displaystyle e^{2x}(4ax^2 +(8a+4b)x +4b +4c +2a) - 2(e^{2x}(2ax^2 +(2a+2b)x +b +2c))$+ $\displaystyle 10(e^{2x}(ax^2+bx + c)) = x^2e^{2x}$

which gives you

$\displaystyle 10a = 1$

$\displaystyle 4a+10b = 0$

$\displaystyle 2a+2b +10c =0$

now solve for a, b, and c and you will have your general solution. I amy have made an alegbra mistake but to sleepy to check right now

8. Originally Posted by pikminman
Not really...

Oh! Yes, I meant to say $\displaystyle \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 10y = 0$ not $\displaystyle \frac{d^2y}{dx^2} - 2\frac{d^2y}{dx^2} + 10y = 0$ Sorry...
Same approach

$\displaystyle r^2 - 2r + 10 = 0$

solve the quadratic equation.

$\displaystyle r^2 -2r = -10$

$\displaystyle (r-1)^2 = -9$

$\displaystyle r-1 = \pm 3i$

$\displaystyle r = 1 \pm 3i$

so you have completed the homo part of the solution

$\displaystyle c_{1}e^x\cos{3x} +c_{2}e^x\sin{3x}$

9. Originally Posted by 11rdc11
Same approach

$\displaystyle r^2 - 2r + 10 = 0$

solve the quadratic equation.

$\displaystyle r^2 -2r = -10$

$\displaystyle (r-1)^2 = -9$

$\displaystyle r-1 = \pm 3i$

$\displaystyle r = 1 \pm 3i$

so you have completed the homo part of the solution

$\displaystyle c_{1}e^x\cos{3x} +c_{2}e^x\sin{3x}$
Hey, could you tell me if the following is even nearly correct?

It's question 1