# 2nd Order Differential Equations

• Feb 16th 2009, 08:46 AM
algorithm
2nd Order Differential Equations
Hello,

Second order differential equation:

y'' - y = exp[x]

Solving for the complementary function, y_c:

m^2 - 1 = 0
-> m = 1 and m = 0

:. y_c = A*exp[-1] + B*exp[0] = A*exp[-1] + B

Solving for the particular integral, y_p:
Try y = a*exp[x] -> this is a solution of the homogeneous eqn, so multiply the function by x:

y = ax*exp[x]
y' = a*exp[x] + ax*exp[x]
y'' = a*exp[x] + a*exp[x] + ax*exp[x]

Substituting into y'' - y' = x*exp[x], the equation reduces to:

a + a*x = x -> How to work the constant (alpha) out?

Thank you
• Feb 16th 2009, 02:27 PM
algorithm
Also, for the differential equation:

y'' - y = Sin(2x)sin(x)

How would one find the particular integral?

Thank you
• Feb 16th 2009, 02:40 PM
Jester
Quote:

Originally Posted by algorithm
Hello,

Second order differential equation:

y'' - y = exp[x]

Solving for the complementary function, y_c:

m^2 - 1 = 0
-> m = 1 and m = 0 (*)

:. y_c = A*exp[-1] + B*exp[0] = A*exp[-1] + B (**)

Solving for the particular integral, y_p:
Try y = a*exp[x] -> this is a solution of the homogeneous eqn, so multiply the function by x:

y = ax*exp[x]
y' = a*exp[x] + ax*exp[x]
y'' = a*exp[x] + a*exp[x] + ax*exp[x]

Substituting into y'' - y' = x*exp[x], (***) the equation reduces to:

a + a*x = x -> How to work the constant (alpha) out?

Thank you

In the step above (*), [tex]m = \pm 1[tex] and in this step (**), it should read

$\displaystyle y_c = A e^x + B e^{-x}$.

And finally in this step (***), you should get from your original ODE

$\displaystyle y'' - y = 2a e^x + ax e^x - ax e^x = 2ae^x = e^x$ giving $\displaystyle a = \frac{1}{2}$ so your solution is

$\displaystyle y = A e^x + B e^{-x} + \frac{1}{2} x e^x$
• Feb 16th 2009, 02:41 PM
Jester
Quote:

Originally Posted by algorithm
Also, for the differential equation:

y'' - y = Sin(2x)sin(x)

How would one find the particular integral?

Thank you

Use the sum/difference formula for $\displaystyle \sin x \sin 2x$.
• Feb 17th 2009, 03:34 PM
algorithm
Hello,

Thank you for the help.

This is another ODE:

$\displaystyle y'' + y' = x^{2}$,
conditions: $\displaystyle y(0) = 0, y'(0) = 0$

Complementary function, y_c:
$\displaystyle m^{2} + m = 0$
$\displaystyle m = 0, -1$

$\displaystyle y_c = (C1 + C2x)e^{-x}$

Particular integral y_p:
Try $\displaystyle y_p = Ax^{2} + Bx + C$
$\displaystyle y_p' = 2Ax + B$
$\displaystyle y_p'' = 2A$

$\displaystyle => 2A + 2Ax + B = x^{2}$
$\displaystyle 2A + B = 0$, and $\displaystyle 2A = 0$, giving $\displaystyle A = B = 0$.

$\displaystyle y = C1 + C2e^{-x}$
$\displaystyle y' = -C2e^{-x}$
$\displaystyle 0 = C1 + C2$
$\displaystyle 0 = -C2$
$\displaystyle C1 = C2 = 0$, giving the final solution:
$\displaystyle y = 0$

The correct answer has more $\displaystyle x$'s.
Thank you
• Feb 17th 2009, 04:01 PM
Jester
Quote:

Originally Posted by algorithm
Hello,

Thank you for the help.

This is another ODE:

$\displaystyle y'' + y' = x^{2}$,
conditions: $\displaystyle y(0) = 0, y'(0) = 0$

Complementary function, y_c:
$\displaystyle m^{2} + m = 0$
$\displaystyle m = 0, -1$

$\displaystyle y_c = (C1 + C2x)e^{-x}$

Particular integral y_p:
Try $\displaystyle y_p = Ax^{2} + Bx + C$ this C is part of your complimentary solution
$\displaystyle y_p' = 2Ax + B$
$\displaystyle y_p'' = 2A$

$\displaystyle => 2A + 2Ax + B = x^{2}$
$\displaystyle 2A + B = 0$, and $\displaystyle 2A = 0$, giving $\displaystyle A = B = 0$.

$\displaystyle y = C1 + C2e^{-x}$
$\displaystyle y' = -C2e^{-x}$
$\displaystyle 0 = C1 + C2$
$\displaystyle 0 = -C2$
$\displaystyle C1 = C2 = 0$, giving the final solution:
$\displaystyle y = 0$

The correct answer has more $\displaystyle x$'s.
Thank you

$\displaystyle y = c_1 + c_2 e^{-x}$
Since part of this complimentary is $\displaystyle x^0$, for your guess you'll need t obump up your particualr to
$\displaystyle y_p = Ax^3 +Bx^2 + Cx$