Inhomogeneous 2nd Order Differential Equation - Particular Integral

Phalanx · 30-05-2009: 30th May 2009 17:23

Hi guys, I'm revising for Differential Equations and I'm having a little bit of trouble with this particular question. It reads:

Consider the following differential equation for :

$y^{ \prime \prime} - 4y = 16e^{2x}$

Find the complementary function for this equation

Find the particular integral for this equation

Write down the solution to this equation, given the initial conditions and $y^{\prime}(0) = 0$ .

I found the complementary function to be:

$y_c (x) = Ae^{2x} + B$

but I have some trouble finding the particular integral.

Since the RHS is in the form $ae^{kx}$ I'm using a trial solution of the form $\alpha e^{kx}$

So I start differentiating

$y_p(x) = \alpha e^{2x}$
${y^\prime}_p (x) = 2 \alpha e^{2x}$
${y^ {\prime \prime}}_p (x) = 4 \alpha e^{2x}$

and then I substitute into:

$y^{ \prime \prime} - 4y = 16e^{2x}$

which gives:

$4 \alpha e^{2x} - 4 \alpha e^{2x} = 16 e^{2x}$

The terms with the alpha coefficients cancel which makes the RHS equal zero.

How do I fix this? Thanks guys.

Necro Defain · 30-05-2009: 30th May 2009 17:26

Originally Posted by Phalanx

Hi guys, I'm revising for Differential Equations and I'm having a little bit of trouble with this particular question. It reads:

Consider the following differential equation for :

$y^{ \prime \prime} - 4y = 16e^{2x}$

Find the complementary function for this equation

Find the particular integral for this equation

Write down the solution to this equation, given the initial conditions and $y^{\prime}(0) = 0$ .

I found the complementary function to be:

$y_c (x) = Ae^{2x} + B$

but I have some trouble finding the particular integral.

Since the RHS is in the form $ae^{kx}$ I'm using a trial solution of the form $\alpha e^{kx}$

So I start differentiating

$y_p(x) = \alpha e^{2x}$
${y^\prime}_p (x) = 2 \alpha e^{2x}$
${y^ {\prime \prime}}_p (x) = 4 \alpha e^{2x}$

and then I substitute into:

$y^{ \prime \prime} - 4y = 16e^{2x}$

which gives:

$4 \alpha e^{2x} - 4 \alpha e^{2x} = 16 e^{2x}$

The terms with the alpha coefficients cancel which makes the RHS equal zero.

How do I fix this? Thanks guys.

Try axe^(2x) if ae^(2x) fails

Phalanx · 30-05-2009: 30th May 2009 17:28

Originally Posted by Necro Defain

Try axe^(2x) if ae^(2x) fails

So that's the way you go about it? Ok, I'll give that a shot.

Thanks

Phalanx · 30-05-2009: 30th May 2009 17:39

Using trial solution of the form $\alpha x e^{2x}$

$y_c(x) = \alpha x e^{2x}$
$y^{\prime}_c(x) = \alpha x 2e^{2x} + \alpha e^{2x} = \alpha e^{2x}(2x + 1)$
$y^{\prime \prime}_c(x) = \alpha x 4e^{2x} + \alpha 2e^{2x} + 2 \alpha e^{2x} = \alpha e^{2x}(4x+4)$

Substituting in

$\alpha e^{2x} (4x+4) - 4 \alpha x e^{2x} = 16e^{2x}$
$16 = \alpha (4x+4) - 4 \alpha x$
$16 = 4 \alpha x + 4 \alpha - 4 \alpha x$
$16 = 4 \alpha$
$\alpha = 4$

Seems right to me. Thanks for the help

rep ++

Necro Defain · 30-05-2009: 30th May 2009 17:48

No problem, thanks for the rep

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Inhomogeneous 2nd Order Differential Equation - Particular Integral