Question

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solve the ordinary differential equation by Laplace Transformation y"-2y-8y=0 if y(0)=3 and y'(0)=6​

Answer 1

Given,

ordinary differential equation: y''-2y'-8y=0.

To Find,

solve the given equation by Laplace transformation if y(0)=3 and y'(0)=6.

Solution,

applying Laplace to the given equation y''-2y'-8y=0.

L[y''-2y'-8y]=0

$s^{2}Y(s)-sy(0)-y'(0)-2sY(s)+2y(0)-8Y(s)=0$

$(s^{2}-2s-8)Y(s)=2s$

$L[y(t)]=2\frac{s}{(s^{2}-2s-8) }$

therefore, $y(t)=3e^{t}cos(3t)+tsint(3t)$

Hence, the Laplace transformation of ordinary differential equation y''-2y'-8y=0 is $y(t)=3e^{t}cos(3t)+tsint(3t)$ .