# Solving 2nd Order Differential Equation: y'' - 4y = 0

• snowJT

## Homework Statement

$$y'' - 4y = 0$$ when y = 1, y' = -1, x = 0

2. The attempt at a solution

$$y'' - 4y = 0$$

$$m^2 - 4 = 0$$

$$m = 2, m = -2$$

Substituting: $$y = 1, x = 0$$

$$1 = C1 + C2$$

$$C1 = -C2 + 1$$

Substituting: $$y' = -1, x = 0$$

$$-1 -C2 + 1 + C2$$

$$C2 = C2$$

Therefore: $$y = -C2e^2^x + C2e^-^2^x$$

Obviously I went wrong with my substitutions... Can I get a hint?

Try solving the initial value as a system of equations, maybe this will give you a better approach than substitution.

since m=+/-2
y is of the form
y=A(e^-2x)+B(e^2x)
y`=-2A(e^-2x)+2B(e^2x)
Use the given conditions. Make two equations... You know the rest.
Hope this helps.
Abdullah

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thanks guys, I got it

## Homework Statement

$$y'' - 4y = 0$$ when y = 1, y' = -1, x = 0

2. The attempt at a solution

$$y'' - 4y = 0$$

$$m^2 - 4 = 0$$

$$m = 2, m = -2$$
It would be a really smart idea to write out the general solution here:
y= C1e2x+ C2e-2x
Otherwise no one knows what the rest of this means!

Substituting: $$y = 1, x = 0$$

$$1 = C1 + C2$$
Okay, C1e0+ C2e0= C1+ C2= 1

$$C1 = -C2 + 1$$

Substituting: $$y' = -1, x = 0$$

$$-1 -C2 + 1 + C2$$
Isn't there an "=" missing here?
If y(x)= C1e2x+ C2e-2x
then y'(x)= 2C1e2x- 2C2e-2x
so y'(0)= 2C1- 2C2= -1

$$C2 = C2$$
Surely that's not what you meant to say!

Therefore: $$y = -C2e^2^x + C2e^-^2^x$$

Obviously I went wrong with my substitutions... Can I get a hint?

You have C1+ C2= 1 and 2C1- 2C2= -1. Solve those two equations.

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