# [SOLVED]17.2.53 Find the unique solution satisfying the equation and the given initial conditions.

#### karush

##### Well-known member
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

Staff member
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

Staff member
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

##### Well-known member
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

Staff member
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.

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

#### HallsofIvy

##### Well-known member
MHB Math Helper
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.