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Math  /  Calculus

QuestionConsider the initial value problem y′′+4y=8t,y(0)=7,y′(0)=4.y^{\prime \prime}+4 y=8 t, \quad y(0)=7, \quad y^{\prime}(0)=4 . a. Take the Laplace transform of both sides of the given differential equation to create the corresponding algebraic equation. Denote the Laplace transform of y(t)y(t) by Y(s)Y(s). Do not move any terms from one side of the equation to the other (until you get to part (b) below). □\square □\square help (formulas) b. Solve your equation for Y(s)Y(s). Y(s)=L{y(t)}=Y(s)=\mathcal{L}\{y(t)\}= □\square c. Take the inverse Laplace transform of both sides of the previous equation to solve for y(t)y(t). y(t)=y(t)= □\square

Studdy Solution

STEP 1

What is this asking? We're given a *second-order differential equation* with *initial conditions* and we need to solve it using the *Laplace transform* method to find y(t)y(t). Watch out! Don't forget to apply the initial conditions correctly when taking the Laplace transform, and double-check the Laplace transform formulas!

STEP 2

1. Laplace Transform of the Differential Equation
2. Solve for Y(s)
3. Inverse Laplace Transform

STEP 3

Let's **apply** the Laplace transform to both sides of the given differential equation: L{y′′+4y}=L{8t} \mathcal{L}\{y'' + 4y\} = \mathcal{L}\{8t\} Remember, the Laplace transform is *linear*, so we can break it up!

STEP 4

L{y′′}+4L{y}=8L{t} \mathcal{L}\{y''\} + 4\mathcal{L}\{y\} = 8\mathcal{L}\{t\} Now, let's use our handy Laplace transform formulas!

STEP 5

We know L{y′′}=s2Y(s)−sy(0)−y′(0)\mathcal{L}\{y''\} = s^2Y(s) - sy(0) - y'(0), L{y}=Y(s)\mathcal{L}\{y\} = Y(s), and L{t}=1s2\mathcal{L}\{t\} = \frac{1}{s^2}. **Substitute** these into our equation: s2Y(s)−sy(0)−y′(0)+4Y(s)=8s2 s^2Y(s) - sy(0) - y'(0) + 4Y(s) = \frac{8}{s^2}

STEP 6

We're given y(0)=7y(0) = 7 and y′(0)=4y'(0) = 4.
Let's **plug** those in: s2Y(s)−7s−4+4Y(s)=8s2 s^2Y(s) - 7s - 4 + 4Y(s) = \frac{8}{s^2}

STEP 7

Let's **factor out** Y(s)Y(s) to **isolate** it: Y(s)(s2+4)−7s−4=8s2 Y(s)(s^2 + 4) - 7s - 4 = \frac{8}{s^2}

STEP 8

Now, **move** the other terms to the right side: Y(s)(s2+4)=8s2+7s+4 Y(s)(s^2 + 4) = \frac{8}{s^2} + 7s + 4

STEP 9

**Divide** both sides by (s2+4)(s^2 + 4) to get Y(s)Y(s) by itself: Y(s)=8s2(s2+4)+7ss2+4+4s2+4 Y(s) = \frac{8}{s^2(s^2 + 4)} + \frac{7s}{s^2 + 4} + \frac{4}{s^2 + 4}

STEP 10

Let's take the **inverse Laplace transform** of both sides: L−1{Y(s)}=L−1{8s2(s2+4)+7ss2+4+4s2+4} \mathcal{L}^{-1}\{Y(s)\} = \mathcal{L}^{-1}\left\{\frac{8}{s^2(s^2 + 4)} + \frac{7s}{s^2 + 4} + \frac{4}{s^2 + 4}\right\}

STEP 11

Using the *linearity* of the inverse Laplace transform, we can **separate** the terms: y(t)=L−1{8s2(s2+4)}+L−1{7ss2+4}+L−1{4s2+4} y(t) = \mathcal{L}^{-1}\left\{\frac{8}{s^2(s^2 + 4)}\right\} + \mathcal{L}^{-1}\left\{\frac{7s}{s^2 + 4}\right\} + \mathcal{L}^{-1}\left\{\frac{4}{s^2 + 4}\right\}

STEP 12

We can **rewrite** the first term using partial fraction decomposition as 2s2−2s2+4\frac{2}{s^2} - \frac{2}{s^2+4}: y(t)=L−1{2s2}−L−1{2s2+4}+L−1{7ss2+4}+L−1{4s2+4} y(t) = \mathcal{L}^{-1}\left\{\frac{2}{s^2}\right\} - \mathcal{L}^{-1}\left\{\frac{2}{s^2 + 4}\right\} + \mathcal{L}^{-1}\left\{\frac{7s}{s^2 + 4}\right\} + \mathcal{L}^{-1}\left\{\frac{4}{s^2 + 4}\right\}

STEP 13

Now, we can **apply** the inverse Laplace transform formulas: y(t)=2t−sin⁡(2t)+7cos⁡(2t)+2sin⁡(2t) y(t) = 2t - \sin(2t) + 7\cos(2t) + 2\sin(2t)

STEP 14

Finally, **combine** like terms: y(t)=2t+sin⁡(2t)+7cos⁡(2t) y(t) = 2t + \sin(2t) + 7\cos(2t)

STEP 15

The solution to the initial value problem is y(t)=2t+sin⁡(2t)+7cos⁡(2t)y(t) = 2t + \sin(2t) + 7\cos(2t).

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