# Thread: 2nd Order Differential Equations

1. ## 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

2. Also, for the differential equation:

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

How would one find the particular integral?

Thank you

3. 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

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

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

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

$y = A e^x + B e^{-x} + \frac{1}{2} x e^x$

4. 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 $\sin x \sin 2x$.

5. Hello,

Thank you for the help.

This is another ODE:

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

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

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

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

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

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

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

6. Originally Posted by algorithm
Hello,

Thank you for the help.

This is another ODE:

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

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

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

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

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

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

The correct answer has more $x$'s.
Thank you
$y = c_1 + c_2 e^{-x}$
Since part of this complimentary is $x^0$, for your guess you'll need t obump up your particualr to
$y_p = Ax^3 +Bx^2 + Cx$