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#### karush

##### Well-known member

- Jan 31, 2012

- 2,928

$y'' + 7y' = 7x$

$y_p=\frac{1}{2}x^2-\frac{1}{7}x$

$y(0)=y'(0)=0$

$r^2+7r=0$

$r=0$

$r=-7$

- Thread starter karush
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- Thread starter
- #1

- Jan 31, 2012

- 2,928

$y'' + 7y' = 7x$

$y_p=\frac{1}{2}x^2-\frac{1}{7}x$

$y(0)=y'(0)=0$

$r^2+7r=0$

$r=0$

$r=-7$

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- #2

This problem is almost identical to the previous. What is your homogeneous solution $y_h$?

$y'' + 7y' = 7x$

$y_p=\frac{1}{2}x^2-\frac{1}{7}x$

$y(0)=y'(0)=0$

$r^2+7r=0$

$r=0$

$r=-7$

- Admin
- #3

\(\displaystyle y''+ay'=bx\) where $y(0)=y_0$ and $y'(0)=y_0'$.

Now, using a technique you'll likely learn in a short while, it can be shown that the particular solution is:

\(\displaystyle y_p(x)=\frac{b}{2a}x^2-\frac{b}{a^2}x\)

From the characteristic equation, we find the homogeneous solution is:

\(\displaystyle y_h(x)=c_1+c_2e^{-ax}\)

And so, by the principle of superposition, we find the general solution to the ODE is:

\(\displaystyle y(x)=y_h(x)+y_p(x)=c_1+c_2e^{-ax}+\frac{b}{2a}x^2-\frac{b}{a^2}x\)

Differentiating, we obtain:

\(\displaystyle y'(x)=-ac_2e^{-ax}+\frac{b}{a}x-\frac{b}{a^2}\)

Now, using our initial values, we find:

\(\displaystyle y(0)=c_1+c_2=y_0\)

\(\displaystyle y'(0)=-ac_2-\frac{b}{a^2}=y_0'\)

This system implies:

\(\displaystyle \left(c_1,c_2\right)=\left(\frac{a^3y_0+a^2y_0'+b}{a^3},-\frac{a^2y_0'+b}{a^3}\right)\)

And so the solution to the IVP is given by:

\(\displaystyle y(x)=\frac{a^3y_0+a^2y_0'+b}{a^3}-\frac{a^2y_0'+b}{a^3}e^{-ax}+\frac{b}{2a}x^2-\frac{b}{a^2}x\)

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- #4

- Jan 31, 2012

- 2,928

I did the very last problem

and got the "fantastic" robot reply

but I followed your example to get it

anyway I need to do more so for this one

$y'' + 7y' = 7x$

$y_p=\frac{1}{2}x^2-\frac{1}{7}x$

$y(0)=y'(0)=0$

$r^2+7r=0$

$r=0$

$r=-7$

$y(x)=c_2\sin(\sqrt{7}x)+c_2\cos(\sqrt{7}x)+x$

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- #5

Only when the roots of the characteristic equation are imaginary do you typically express the homogeneous solution as a sinusoid. For example, if the given ODE had been:

I did the very last problem

and got the "fantastic" robot reply

but I followed your example to get it

anyway I need to do more so for this one

$y'' + 7y' = 7x$

$y_p=\frac{1}{2}x^2-\frac{1}{7}x$

$y(0)=y'(0)=0$

$r^2+7r=0$

$r=0$

$r=-7$

$y(x)=c_2\sin(\sqrt{7}x)+c_2\cos(\sqrt{7}x)+x$

\(\displaystyle y''+7y=f(x)\)

Then, the characteristic equation would be:

\(\displaystyle r^2+7=0\implies r=\pm\sqrt{7}i\)

And the homogeneous solution would then be:

\(\displaystyle y_h(x)=c_1\cos(\sqrt{7}x)+c_2\sin(\sqrt{7}x)\)

But in this problem, and the one before, the roots are real, and so we would use the exponential form for our homogeneous solution.

The other case, when the roots are real and repeated, will be a discussion for another thread.

- Jan 29, 2012

- 1,151