-17.2.53 Find the unique solution from initial conditions.

[karush]

the function y_p is a particular solution to the specified nonhomogeneous equation. Find the unique solution satisfying the equation and the given initial conditions.
$y'' + 7y' = 7x$
$y_p=\frac{1}{2}x^2-\frac{1}{7}x$
$y(0)=y'(0)=0$
$r^2+7r=0$
$r=0$
$r=-7$

 

[MarkFL]

the function y_p is a particular solution to the specified nonhomogeneous equation. Find the unique solution satisfying the equation and the given initial conditions.
$y'' + 7y' = 7x$
$y_p=\frac{1}{2}x^2-\frac{1}{7}x$
$y(0)=y'(0)=0$
$r^2+7r=0$
$r=0$
$r=-7$

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

 

[MarkFL]

What I used to like to do as a student, when presented with a batch of similar problems, is to generalize to get a formula I could use. Suppose we are given the IVP:

\(\displaystyle y''+ay'=bx\) 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:

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

 

[karush]

wow thank you

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
$y'' + 7y' = 7x$
$y_p=\frac{1}{2}x^2-\frac{1}{7}x$
$y(0)=y'(0)=0$
$r^2+7r=0$
$r=0$
$r=-7$
$y(x)=c_2\sin(\sqrt{7}x)+c_2\cos(\sqrt{7}x)+x$

 

[MarkFL]

wow thank you

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
$y'' + 7y' = 7x$
$y_p=\frac{1}{2}x^2-\frac{1}{7}x$
$y(0)=y'(0)=0$
$r^2+7r=0$
$r=0$
$r=-7$
$y(x)=c_2\sin(\sqrt{7}x)+c_2\cos(\sqrt{7}x)+x$

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:

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

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. :D

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

 

[HallsofIvy]

With characteristic roots 0 and -7, the general solution to the homogeneous differential equation is \(\displaystyle Ce^{0x}+ De^{-7x}= C+ De^{-7x}\). To solve y''+ 7y'= 7x, look for a specific solution of the form \(\displaystyle y= (Ax+ B)x= Ax^2+ Bx\). Then y'= 2Ax+ B and y''= 2A so the equation becomes 2A+ 14Ax+ 7B= 7x. We must have 14A= 7 and 2A+ 7B= 0.