# 2nd order differential

• August 9th 2008, 12:18 AM
divine_mail
2nd order differential
can anyone give msom light on this question, thanks

find the complete solution of the differential equation
d2y/dx2 – 4 dy/dx + 3y = 65 sin 2 x

find the value of the constant a such that y= axe^-1 is a solution of the differential equation
d2y/dx2 + 3 dy/dx + 2y = 2 e-x
find the solution for which y=1 and dy/dx =3 when x=0
• August 9th 2008, 03:15 AM
mr fantastic
Quote:

Originally Posted by divine_mail
can anyone give msom light on this question, thanks

find the complete solution of the differential equation
d2y/dx2 – 4 dy/dx + 3y = 65 sin 2 x

Mr F says: It's expected that you can find the homogeneous solution, that is, the solution to ${\color{red}\frac{d^2 y}{dx^2} - 4 \frac{dy}{dx} + 3y = 0}$.

So I'll assume that your trouble is in getting a particular solution.

It's unclear from your notation whether you mean ${\color{red}65 \sin (2x)}$ or ${\color{red}65 \sin^2 x}$.

If ${\color{red}65 \sin (2x)}$ then try ${\color{red} y = a \sin (2x) + b \cos (2x)}$ as a particular solution.

If ${\color{red}65 \sin^2 x}$ then note that ${\color{red}65 \sin^2 x = \frac{65}{2} (1 - \cos (2x)) = \frac{65}{2} - \frac{65}{2} \cos (2x)}$ and so you should try ${\color{red} y = a \sin (2x) + b \cos (2x) + c}$ as a particular solution.

In each case you substitute the particular solution into the DE and solve for the values of a, b, c .......

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find the value of the constant a such that y= axe^-1 is a solution of the differential equation Mr F says: You mean y= axe^-x.

d2y/dx2 + 3 dy/dx + 2y = 2 e-x

Mr F says: You're being asked this question because it's the form of a particular solution to the DE.

The general solution to the DE is the homogeneous solution plus the particular solution.

It's expected that you can find the homogeneous solution. It's ${\color{red} y = A e^{-x} + B e^{-2x}}$.

Then the general solution is ${\color{red} y = A e^{-x} + B e^{-2x} + a x e^{-x}}$. The value of a is found below.

find the solution for which y=1 and dy/dx =3 when x=0

Mr F says: Substitute these boundary conditions into the general solution to get the value of A and B.

$y = a x e^{-x}$ therefore:

$\frac{dy}{dx} = a e^{-x} - ax e^{-x} = a e^{-x} (1 - x)$.

$\frac{d^2y}{dx^2} = - a e^{-x} (1 - x) - a e^{-x} = -a e^{-x} (2 - x)$.

Substitute these expressions into the DE and simplify:

$-a e^{-x} (2 - x) + 3 a e^{-x} (1 - x) + 2 a x e^{-x} = 2 e^{-x}$

$\Rightarrow -a (2 - x) + 3a (1 - x) + 2ax = 2$

$\Rightarrow a = 2$.