solve the laplace transform y\'\'+y=cos2t, y(0)=0, y\'(0)=1 Solution First take

Question

Explanation / Answer

First take the Laplace transform of each term and apply the initial conditions. L(y") + w²L(y) = L(cos 2t) L(y") = s²Y(s) - sy(0) - y'(0) = s²Y(s) - s L(y) = Y(s) L(cos 2t) = s/(s² + 2²) = s/(s² + 4) s²Y(s) - s + w²Y(s) = s/(s² + 4) Solve for Y(s) (s² + w²)Y(s) = s + s/(s² + 4) Y(s) = [s + s/(s² + 4)]/(s² + w²) Y(s) = s/(s² + w²) + s/[(s² + w²)(s² + 4)] Since you want y(t), you can just take the inverse LT of Y(s). In the current form you'll have to do partial fraction decomp twice, so add the fractions into one term. Combining terms so we have a fraction with a single denominator, Y(s) = [s(s² + 4) + s]/[(s² + w²)(s² + 4)] = (s³ + 5s)/[(s² + w²)(s² + 4)] Perform partial fraction decomposition. Y(s) = (as + b)/(s² + w²) + (cs + d)/(s² + 4) = [(as + b)(s² + 4) + (cs + d)(s² + w²)]/[(s² + w²)(s² + 4)] = (as³ + 4as + bs² + 4b + cs³ + cw²s + ds² + dw²)/[(s² + w²)(s² + 4)] = [(a + c)s³ + (b + d)s² + (4a + cw²) + 4b + dw²]/[(s² + w²)(s² + 4)] Equating numerators: s³ + 5s = (a + c)s³ + (b + d)s² + (4a + cw²)s + 4b + dw² Setting coefficients equal: s³: 1 = a + c s: 5 = 4a + cw² 5 = 4(1 - c) + cw² 5 = 4 - 4c + cw² c = 1/(w² - 4) ---> Now we know why w² ? 4 restriction is in place. a = 1 - 1/(w² - 4) s²: 0 = b + d constant: 0 = 4b + dw² 0 = -4d + dw² d = b = 0 Y(s) = [1 - 1/(w² - 4)]/(s² + w²) + [1/(w² - 4)]/(s² + 4) Now take the inverse LT. L?¹[Y(s)] = L?¹[1 - 1/(w² - 4)]/(s² + w²)] = L?¹[1/(w² - 4)]/(s² + 4) L?¹[Y(s)] = y(t) L?¹[1 - 1/(w² - 4)]/(s² + w²)] = [1 - 1/(w² - 4)] L?¹[1/(s² + w²)] = {[1 - 1/(w² - 4)]sin(wt)}/w L?¹[1/(w² - 4)]/(s² + 4) = [1/(w² - 4)] L?¹[1/(s² + 4)] = ½[1/(w² - 4)]sin(2t) y(t) = {[1 - 1/(w² - 4)]sin(wt)}/w + ½[1/(w² - 4)]sin(2t) = sin(wt)/w - sin(wt)/[w(w² - 4) + sin(2t)/[2(w² - 4) Combine to one fraction with denominator 2w(w² - 4). y(t) = [2(w² - 4)sin(wt) - 2sin(wt) + w sin(2t)]/[2w(w² - 4)] = [(2w² - 10)sin(wt) + w sin(2t)]/[2w(w² - 4)] = [(w² - 5)sin(wt) + sin(2t)]/(w² - 4)

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.