2nd Order differential equations

**ashura** · November 15th 2006, 06:29 AM

Need help to complete this question,

$\frac{d^2 y}{dx^2}-5\frac{dy}{dx}+6y = x^2$

$D^2-5D+6 = 0$

$D^2-3D-2D+6 = 0$

$D(D-3) -2(D-3) = 0$

$(D-2)(D-3) = 0$

$D = 2,3$

$\therefore y_c = Ae^{2x}+Be^{3x}$

$(D^2-5D+6)y_p = x^2$

$y_p = \frac{x^2}{(D-2)(D-3)}$

...................................
...................................

**ThePerfectHacker** · November 15th 2006, 07:53 AM

Originally Posted by ashura

Need help to complete this question,

$\frac{d^2 y}{dx^2}-5\frac{dy}{dx}+6y = x^2$

$D^2-5D+6 = 0$

$D^2-3D-2D+6 = 0$

$D(D-3) -2(D-3) = 0$

$(D-2)(D-3) = 0$

$D = 2,3$

$\therefore y_c = Ae^{2x}+Be^{3x}$

$(D^2-5D+6)y_p = x^2$

$y_p = \frac{x^2}{(D-2)(D-3)}$

...................................
...................................

I have seen that operator before but not sure how it works.
Just look for a solution of the form $Ax^2+Bx+C$ by undetermined coefficients.

Thus,
$6y=6Ax^2+6Bx+6C$
$-5y'=-10Ax-5B$
$y''=2A$
Thus, adding
$6Ax^2+(6B-10A)x+(6C-5B+2A)=x^2$
So solve,
$6A=1$
$6B-10A=0$
$6C-5B+2A=0$

And your general solution is correct!

**Soroban** · November 15th 2006, 12:15 PM

Hello, ashura!

$\frac{d^2 y}{dx^2}-5\frac{dy}{dx}+6y \:= \:x^2$

$D^2-5D+6 \:= \:0\quad\Rightarrow\quad (D-2)(D-3) \:= \:0\quad\Rightarrow\quad D \,= \,2,\,3$

. . $\therefore \;y_c \:= \:Ae^{2x}+Be^{3x}$

$(D^2-5D+6)y_p \:= \:x^2\quad\Rightarrow\quad y_p \:= \:\frac{x^2}{(D-2)(D-3)}$

Evidently, you're using the Method of Operators.

You're expected to know that: . $\boxed{\frac{1}{D - a}\!\cdot\!f(x) \:=\:e^{ax}\int e^{-ax}f(x)\,dx}$

Hence: . $\frac{1}{D-3}(x^2)\;=\;e^{3x}\underbrace{\int e^{-3x}x^2\,dx}_{\text{integrate by parts}}$

. . $= \;e^{3x}\left[-\frac{1}{3}x^2e^{-3x} - \frac{2}{9}xe^{-3x} - \frac{2}{27}e^{-3x} + C_1\right]$

. . $= \;-\frac{1}{3}x^2 - \frac{2}{9}x - \frac{2}{27} + C_1e^{3x} \;= \;-\frac{1}{27}\left(9x^2 + 6x + 2 + C_1e^{3x}\right)$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I'll let you take it through the second stage . . .

$\frac{1}{D - 2}\left[-\frac{1}{27}\left(9x^2 + 6x + 2 + C_1e^{3x}\right)\right]$

. . . $= \;e^{2x}\int e^{-2x}\left[-\frac{1}{27}\left(9x^2 + 6x + 2 + C_1e^{3x}\right)\right]\,dx$

. . . $= \;-\frac{1}{27}e^{2x}\int \left[\left(9x^2 + 6x + 2\right)e^{-2x} + C_1e^x\right]\,dx$ . . . etc.

**ashura** · November 16th 2006, 03:06 AM

Thanks for your help. I can now complete the solution.

Thread: 2nd Order differential equations

LinkBack

Thread Tools

Display

2nd Order differential equations

Similar Math Help Forum Discussions

Decomposition of 2 fourth order differential equations on 4 equations of second order

Differential Equations-Application of 1st order differential equations 1

Higher-order differential equations, the Laplace transform and exponential order

Differential Equations:Linear first Order Equations

First Order Differential Equations: Separable Equations and Applications

Search Tags