# Thread: Two Webwork Questions (Second order)

1. ## Two Webwork Questions (Second order)

I've been attempting these two questions a few times and I countinue to get them wrong.

1) Find y as a function of x if
$\displaystyle x^2(y'') -15xy' + 64y = x^3$ y(1)=6, y'(1)=-5
So I solved the Homogenous equation as a Euler equation. My homogeneous solution is yh= C1x^(8)+C2ln(x)x^(8)
then i proceeded with undetermined coefficients and ended up with 151/25x^8-1336/25ln((x))x^8+x^(3)/25
Whats wrong with the method?

2) Use the method of undetermined coefficients or the method of differential operators to find one solution of
y'' -8y' + 43y = 32e^4t(cos(5t))+64e^4t(sin(5t))+ 7e^0t
(It doesn't matter which specific solution you find for this problem.)
My guess for yp was Ae^4t cos(5t)+Be^ 4t sin(5t)
And I used the undetermined coefficent method to find A=16 and B=32.

Thanks.

2. As to (1), what did you get as your specific solution to the non-homogeneous equation (to which you added the general homogeneous solution)?

--Kevin C.

3. for (1) my specific solution was y=x^(3)/25

4. anyone have any idea whats wrong ?

5. Your specific solution seems to be correct, as does your general homgeneous solution. Thus, you have general solution $\displaystyle y(x)=(C_1+C_2\ln{x})x^8+\frac{x^3}{25}$.
Setting y(1)=6, we get
$\displaystyle 6=(C_1+C_2\ln(1))1^8+\frac{1^3}{25}$
$\displaystyle 6=(C_1+C_2\cdot0)+\frac{1}{25}$
$\displaystyle 6=C_1+\frac{1}{25}$
$\displaystyle C_1=6-\frac{1}{25}=\frac{149}{25}$
you have $\displaystyle C_1=\frac{151}{25}=6+\frac{1}{25}$, which by the above, you can see is incorrect.
With $\displaystyle C_1=\frac{149}{25}$, set y'(1)=-5, and solve for the correct $\displaystyle C_2$.

--Kevin C.

6. As to (2),note that $\displaystyle 7e^{0t}=7e^{0}=7$, so that your differential equation is
$\displaystyle y''-8y'+43y=32e^{4t}\cos{5t}+64e^{4t}\sin{5t}+7e^{0t}$
which is
$\displaystyle y''-8y'+43y=32e^{4t}\cos{5t}+64e^{4t}\sin{5t}+7$
what you gave as a solution is in fact a solution to
$\displaystyle y''-8y'+43y=32e^{4t}\cos{5t}+64e^{4t}\sin{5t}$,
not the above equation.
[Hint: consider what happens when you add a constant to the solution you have].

--Kevin C.