# Math Help - Differential equation and its y(t) answer given a u(t) statement

1. ## Differential equation and its y(t) answer given a u(t) statement

Given the differential equation:
$\frac{d^{4}y}{dt^{4}}+6\frac{d^{2}y}{dt^{2}}+25y=u '+3u$

Find the $y(t)$ solution when $u(t)=e^{t}\cos t$
Hint: $e^{t}\cos t=Re(e^{(1+i)t})$

$y(t)=\frac{32}{195}e^{t}\cos t+\frac{9}{195}e^{t}\sin t$

My attempt (I'm not sure if I'm using the correct method):
I start by using the "guessing method" which states that if you have $u(t)=sin(kt)$ then you can guess on the solution with the form:
$y(t)=A \cos(kt)+B \sin(kt)$

So my guess so far is:
$y(t)=A \cos(t)+B \sin(t)$

I hereby differentiate this guess 2 and 4 times and insert it to my original equation:

$y''(t)=-A \cos(t)-B \sin(t)$
$y''''(t)=A \cos(t)+B \sin(t)$

Inserted:
$(A \cos(t)+B \sin(t))+6\cdot (-A \cos(t)-B \sin(t))+25\cdot (A \cos(t)+B \sin(t))=u'+3u$

I clean it up a little:
$20\cdot A \cos(t)+20\cdot B \sin(t)=u'+3u$

This is as far as I've made it because in the examples in my textbook, they have not shown what to do when $u$ appears in the original equation, thus I have no idea how to isolate and find the values of $A$ and $B$.

Can somebody please tell me if I'm on right track at all, and if so, what do I do next?

2. ## Re: Differential equation and its y(t) answer given a u(t) statement

Hey MathIsSoHard.

This is a linear ordinary differential equation (since you have the explicit form for u(t)).

Try using Differential Operator techniques. Let P(D) = (D^4 + 6D^2 + 25). You will need to solve 1/P(D)[u' + 3u] for the specific solution p_s and solve P(D)y = 0 using the substitution y = e^(mx) for the complementary solution p_c.

Then your solution will be the sum of these giving y(t) = p_c + p_s (which are both functions of t).

3. ## Re: Differential equation and its y(t) answer given a u(t) statement

Originally Posted by chiro
Hey MathIsSoHard.

This is a linear ordinary differential equation (since you have the explicit form for u(t)).

Try using Differential Operator techniques. Let P(D) = (D^4 + 6D^2 + 25). You will need to solve 1/P(D)[u' + 3u] for the specific solution p_s and solve P(D)y = 0 using the substitution y = e^(mx) for the complementary solution p_c.

Then your solution will be the sum of these giving y(t) = p_c + p_s (which are both functions of t).
I've worked a little more on it, this is how far I've made it so far:

Given the differential equation:
$\frac{d^{4}y}{dt^{4}}+6\frac{d^{2}y}{dt^{2}}+25y=u '+3u$

And we have that $u(t)=e^{t}\cos(t)$

If I substitute $u(t)=e^{t}\cos t$ on the right side of the equation, I get:
$\frac{d^{4}y}{dt^{4}}+6\frac{d^{2}y}{dt^{2}}+25y=e ^t \cdot \cos(t) - e^t \cdot \sin(t) + 3 \cdot e^t \cdot \cos(t)$

I start by using the "guessing method" which states that if you have $u(t)=sin(kt)$ then you can guess on the solution with the form:
$y(t)=A \cos(kt)+B \sin(kt)$
And if you have $u(t)=e^{st}$

So my guess so far is:
$y(t)=A\cdot e^t \cdot \cos(t)+B \cdot e^t \cdot \sin(t)$

I hereby differentiate this guess 2 and 4 times and insert it to my original equation:

$y''(t)=-2 \cdot A \cdot e^t \cdot \sin(t) + 2 \cdot B \cdot e^t \cdot \cos(t)$
$y''''(t)=-4 \cdot A \cdot e^t \cdot \cos(t) - 4 \cdot B \cdot e^t \cdot \sin(t)$

Inserted into the original equation:
$-4 \cdot A \cdot e^t \cdot \cos(t) - 4 \cdot B \cdot e^t \cdot \sin(t)+6\cdot (-2 \cdot A \cdot e^t \cdot \sin(t) + 2 \cdot B \cdot e^t \cdot \cos(t))=-e^t \cdot \sin(t) + 4 \cdot e^t \cdot \cos(t)$

I clean it up a little:
$(-4A+12B+25A)\cdot e^t \cdot \cos(t) + (-4B-12A+25B)\cdot e^t \sin(t)=-e^t \cdot \sin(t) + 4 \cdot e^t \cdot \cos(t)$

The coefficients of the trigonometric functions is:
$-4A+12B+25A=4$
$-4B-12A+25B=-1$

Two equations with two unknown factors (A and B)
and I get the values to:
$A=\frac{32}{195}$
$B=\frac{3}{65}$

$y(t)=\frac{32}{195}e^{t}\cos t+\frac{9}{195}e^{t}\sin t$
So my A value is correct, but my B value should be $\frac{9}{195}$???
$\frac{3}{65}$ is the same as $\frac{9}{195}$