Studdy AI Logo
Math  /  Calculus

Question2) Solve the initial-value problem 2y′′+y=4δ(t−π);y(0)=0,y′(0)=12 y^{\prime \prime}+y=4 \delta(t-\pi) ; \quad y(0)=0, y^{\prime}(0)=1
HINT: Use the example we discussed in class from section 6.5 as a guide.

Studdy Solution

STEP 1

What is this asking? We need to find a function y(t)y(t) that satisfies the given differential equation 2y′′+y=4δ(t−π)2y'' + y = 4\delta(t - \pi) with the **initial conditions** y(0)=0y(0) = 0 and y′(0)=1y'(0) = 1.
This involves a **Dirac delta function**, so we're dealing with an impulse at time t=πt = \pi. Watch out! Don't forget to apply the initial conditions *after* taking the Laplace transform!
Also, remember the special properties of the Laplace transform of the Dirac delta function.

STEP 2

1. Laplace Transform
2. Partial Fractions
3. Inverse Laplace Transform

STEP 3

Let's **kick things off** by taking the Laplace transform of both sides of the differential equation.
Remember, the Laplace transform is a powerful tool that turns our differential equation into an algebraic equation!
We get: L{2y′′+y}=L{4δ(t−π)} \mathcal{L}\{2y'' + y\} = \mathcal{L}\{4\delta(t - \pi)\}

STEP 4

Using the linearity property of the Laplace transform and remembering our initial conditions, this becomes: 2[s2Y(s)−sy(0)−y′(0)]+Y(s)=4e−πs 2[s^2Y(s) - sy(0) - y'(0)] + Y(s) = 4e^{-\pi s} Substituting y(0)=0y(0) = 0 and y′(0)=1y'(0) = 1, we get: 2[s2Y(s)−0−1]+Y(s)=4e−πs 2[s^2Y(s) - 0 - 1] + Y(s) = 4e^{-\pi s} 2s2Y(s)−2+Y(s)=4e−πs 2s^2Y(s) - 2 + Y(s) = 4e^{-\pi s}

STEP 5

Now, let's **isolate** Y(s)Y(s): (2s2+1)Y(s)=2+4e−πs (2s^2 + 1)Y(s) = 2 + 4e^{-\pi s} Y(s)=22s2+1+4e−πs2s2+1 Y(s) = \frac{2}{2s^2 + 1} + \frac{4e^{-\pi s}}{2s^2 + 1} Y(s)=1s2+12+2e−πss2+12 Y(s) = \frac{1}{s^2 + \frac{1}{2}} + \frac{2e^{-\pi s}}{s^2 + \frac{1}{2}}

STEP 6

Good news, everyone!
No partial fractions are needed in this problem!
We already have expressions that look like something we can handle with the inverse Laplace transform.
Let's **rewrite** the expression slightly to make it even clearer: Y(s)=1s2+(12)2+2e−πs1s2+(12)2 Y(s) = \frac{1}{s^2 + (\frac{1}{\sqrt{2}})^2} + 2e^{-\pi s}\frac{1}{s^2 + (\frac{1}{\sqrt{2}})^2}

STEP 7

Time to bring it home!
We'll take the inverse Laplace transform of both sides of our equation.
Remember, the inverse Laplace transform takes us back from the ss-domain to the tt-domain, giving us our solution y(t)y(t)! y(t)=L−1{Y(s)} y(t) = \mathcal{L}^{-1}\{Y(s)\}

STEP 8

Recall that L−1{1s2+a2}=1asin⁡(at)\mathcal{L}^{-1}\{\frac{1}{s^2 + a^2}\} = \frac{1}{a}\sin(at) and L−1{e−csF(s)}=uc(t)f(t−c)\mathcal{L}^{-1}\{e^{-cs}F(s)\} = u_c(t)f(t-c).
Using these, we get: y(t)=2sin⁡(t2)+2uπ(t)2sin⁡(t−π2) y(t) = \sqrt{2}\sin(\frac{t}{\sqrt{2}}) + 2u_\pi(t)\sqrt{2}\sin(\frac{t-\pi}{\sqrt{2}})

STEP 9

Our solution to the initial-value problem is: y(t)=2sin⁡(t2)+22uπ(t)sin⁡(t−π2) y(t) = \sqrt{2}\sin(\frac{t}{\sqrt{2}}) + 2\sqrt{2}u_\pi(t)\sin(\frac{t-\pi}{\sqrt{2}})

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord