Question

Take Laplace Transformation on the Ordinary Differential Equation if y’’’ – 3y’’ + 3y’ – y = t^2 e^t if y(0) = 1, y’(0) = b and y’’(0) = c.

(a) \((s^3-3s^2+3s-1)Y(s)+(-as^2+(3a-b)s+(-3a-c))=\frac{2}{(s-1)^3} \)

(b) \((s^3-3s^2+3s-1)Y(s)+(-as^2+(3a-b)+(-3a-c)s)=\frac{2}{(s-1)^3} \)

(c) \((s^3-3s^2+3s)Y(s)+(-as+(3a-b)s+(-3a-c))=\frac{2}{(s-1)^3} \)

(d) \((s^3-3s^2+3s-1)Y(s)+(-as^2+(3a-b)s+(-3a-c))=\frac{2}{(s-1)^3} \)

This question was posed to me in class test.

My question is from Solution of DE With Constant Coefficients using the Laplace Transform in division Laplace Transform of Engineering Mathematics

Answer 1

The correct option is (a) \((s^3-3s^2+3s-1)Y(s)+(-as^2+(3a-b)s+(-3a-c))=\frac{2}{(s-1)^3} \)

Explanation: L[y’’’ – 3y’’ + 3y’ – y = t^2 e^t]

s^3 Y(s) – s^2 y(0) – sy'(0) – y”(0) – 3s^2 Y(s) + 3sy(0) + 3y'(0) + 3sY(s) – 3y(0) – Y(s) = \(\frac{2}{(s-1)^3} \)

\((s^3-3s^2+3s-1)Y(s)+(-as^2+(3a-b)s+(-3a-c))=\frac{2}{(s-1)^3}.\)