Question

using the laplace transform find the solution of y"+25y=10 where y(0) =2, y'(0)=0

Answer 1

Answer:

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Answer 2

The solution of the differential equation by Laplace transform is $y = \dfrac{2}{5} (1+4cos5t)$

Step-by-step explanation:

Given: Differential equation $y''+25y=10$ and $y(0) = 2$ & $y'(0)=0$

To Find: Solution of the differential equation by Laplace transform

Solution:

• Solving the given differential equation by Laplace transform

We have the differential equation $y''+25y=10$. Taking Laplace both sides, we get,

$L[y'']+25L[y]=10L[1]$

$\Rightarrow (s^2L[y] - sy(0)-y'(0))+25L[y]= \dfrac{10}{s}\\ {}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \because L[y''] = s^2L[y] - sy(0)-y'(0) \ \& \ L[1] = \dfrac{1}{s}$

$\Rightarrow (s^2+25)L[y] = 2s + \dfrac{10}{s}$

$\Rightarrow L[y] = \dfrac{2s}{(s^2+25)} + \dfrac{10}{s(s^2+25)}$

By using the partial fraction method, we rewrite the 2nd term as,

$\dfrac{10}{s(s^2+25)} = \dfrac{2}{5} \Big(\dfrac{1}{s}\Big)- \dfrac{2s}{5} \Big(\dfrac{1}{s^2+25}\Big)$

Therefore,

$\Rightarrow L[y] = \dfrac{2s}{(s^2+25)} + \dfrac{2}{5} \Big(\dfrac{1}{s}\Big)- \dfrac{2s}{5} \Big(\dfrac{1}{s^2+25}\Big)$

Now, taking Laplace inverse both sides, we get

$\Rightarrow y = 2 L^{-1} \Big[\dfrac{s}{(s^2+25)} \Big] + \dfrac{2}{5} L^{-1}\Big[\dfrac{1}{s}\Big]- \dfrac{2}{5} L^{-1}\Big[\dfrac{s}{s^2+25}\Big]$

$\Rightarrow y = 2 cos5t + \dfrac{2}{5} - \dfrac{2}{5}cos5t \ \because L^{-1}\dfrac{s}{s^2 + a^2} = cosat \ \& \ L^{-1}\dfrac{1}{s} = 1$

$\Rightarrow y = \dfrac{2}{5} (1+4cos5t)$

Hence, the solution of the differential equation by Laplace transform is $y = \dfrac{2}{5} (1+4cos5t)$