General solution of reduction of order

sumrtym77

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General solution of y" + 8y + q(x) = 0 y=e^(-4x) I know y'=0y'
I don't know what to do with the q(x). I know what to do without the q(x) so could somebody tell me what to with it? Also, should I have _+_+_ or _+_ 3 solutions or two solutions?
 
sumrtym77 said:
General solution of y" + 8y + q(x) = 0 y=e^(-4x) I know y'=0y'

I do not know what you are doing.



I don't know what to do with the q(x). I know what to do without the q(x) so could somebody tell me what to with it? Also, should I have _+_+_ or _+_ 3 solutions or two solutions?

general solution is:

y(x) = y[sub:36g0yfwm]h[/sub:36g0yfwm](x) + y[sub:36g0yfwm]p[/sub:36g0yfwm](x) <<<< Homogeneous equation + Particular equation

to find homogeneous solution solve:

y" + 8y = 0

particular solution depends on the nature of q(x).
 
One solution (the equation I gave) is y1= e^(-4x). Use the reduction-of-order method to determine a second, linearly independent solution.
 
sumrtym77 said:
One solution (the equation I gave) is y1= e^(-4x)

Was this solution given to you - or calculated it?
.
Use the reduction-of-order method to determine a second, linearly independent solution.
 
Does this look correct?

y1=e^(-4x)
y2=c1y1 *[Se^(-Ssdx) dx]
s= 2 (y1'/y1) + P P=y' , so P=0
s= 2 (-4e(-4x))/e^(-4x) so, s=-8

y2= c1e^(-4x)*[Se^(-S -8 dx) dx]
y2= c1e^-4x * [ e^(8x) / 8]
y2= c1e^(4x)

So, y(x)= A e^(4x)/8 + Be^(-4x)

Is that right?
 
sumrtym77 said:
Does this look correct?

y1=e^(-4x)
y2=c1y1 *[Se^(-Ssdx) dx]
s= 2 (y1'/y1) + P P=y' , so P=0
s= 2 (-4e(-4x))/e^(-4x) so, s=-8

y2= c1e^(-4x)*[Se^(-S -8 dx) dx]
y2= c1e^-4x * [ e^(8x) / 8]
y2= c1e^(4x)

So, y(x)= A e^(4x)/8 + Be^(-4x) <<< Check and see if it satisfies the original ODE.

Is that right?
 
Here is a general formula when you are given two solutions \(\displaystyle y_{1}(t), \;\ y_{2}(t)\)

If you have a second order equation, \(\displaystyle y''+a_{1}(t)y'+a_{2}y=b(t)\)

We have \(\displaystyle W_{1}(t)=-y_{2}(t), \;\ W_{2}(t)=y_{1}(t)\) for any pair of fundamental pair of solutions \(\displaystyle y_{1}(t), \;\ y_{2}(t)\) of a

homogeneous equation. If W(t) is their Wronskian, then the solution y(t) satisfying \(\displaystyle t(t_{0})=y_{0}, \;\ y'(t_{0})=y'_{0}\) is given

by, brace yourself, \(\displaystyle y(t)=c_{1}y_{1}(t)+c_{2}y_{2}(t)-y_{1}(t)\int_{t_{0}}^{t}\frac{b(s)y_{2}(s)}{W(s)}ds+y_{2}(t)\int_{t}^{t_{0}}\frac{b(s)y_{1}(s)}{W(s)}ds\). s is a dummy variable, so to speak.

Where \(\displaystyle c_{1}, \;\ c_{2}\) are constants chosen so that \(\displaystyle {\phi}(t)=c_{1}y_{1}(t)+c_{2}y_{2}(t)\) satisfies the initial conditions.

I know....i know...it looks monstrous. But it is not as bad as it looks. You are given the info to solve. It is a general form to find the other solution.

For instance, suppose we have \(\displaystyle y_{1}(t)=\sqrt{t}, \;\ y_{2}(t)=\sqrt{t}ln(t)\) are a solution of the fundamental system of solutions of the equation

\(\displaystyle y''+\frac{1}{4t^{2}}y=0\) and W(t)=1.

Using the expression and letting \(\displaystyle t_{0}=1\), we can find out that the solution y(t) satisfying y(1)=1 and y'(1)=3/2 of

\(\displaystyle y''+\frac{1}{4t^{2}}y=t^{\frac{3}{2}}\) is given by \(\displaystyle y(t)=\frac{8}{9}\sqrt{t}+\frac{2}{3}\sqrt{t}ln(t)+\frac{1}{9}t^{\frac{7}{2}}\)

This is what I use to solve many reduction of order problems. It can make them easier.
 
There is something wrong with the problem:

y" + 8y + q(x) = 0 y = e^(-4x)

The given equation does not satisfy the homogeneous part of the ODE.

y' = -4 * e^(-4x)

y" = 16*e^(-4x)

This does not satisfy the homogeneous equation. The only way this can be part of the solution is if

q(x) = 16*e^(-4x)

The homogeneous solution of the given ODE is sinosoidal.
 
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