Finding solution of ODE using Laplace transformwith initial condition y(0)=0.y′+2y=f(x) where f(x)=0 if x&gt;1 and f(x)=1 if 0≤x≤1.I applied Laplace transform on both sides which getssY(s)+2Y(s)=1−e−ss⇒Y(s)=1−e−ss(s+2)=1s(s+2)−e−ss(s+2)Now I know inverse Laplace transform of 1s(s+2) but how to find inverse Laplace transform of second term? Or can we solve above ODE using different method?

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Finding solution of ODE using Laplace transformwith initial condition y(0)=0.y′+2y=f(x) where f(x)=0 if x&amp;gt;1 and f(x)=1 if 0≤x≤1.I applied Laplace transform on both sides which getssY(s)+2Y(s)=1−e−ss⇒Y(s)=1−e−ss(s+2)=1s(s+2)−e−ss(s+2)Now I know inverse Laplace transform of 1s(s+2) but how to find inverse Laplace transform of second term? Or can we solve above ODE using different method?

Mey9ci0 · Accepted Answer

Step 1The Laplace transform is a little bit of an overkill for tis equation. Variation of parameters does the trick:The general solution of your equation is:&amp;nbsp;y(x)=c⋅e−2x+e−2x∫0xe2sf(s)dsStep 2Using your intial data and&amp;nbsp;f&amp;nbsp;(it is not clearly defined) we end up withy(x)=e−2x∫0min(x,1)e2sds&amp;nbsp;=e−2xe2min(x,1)−12&amp;nbsp;=e2min(0,1−x)−e−2x2That being said of course you equation can be solved with Laplace transform. You could handle the second term with a partial fraction decomposition and than use a table of standard transformations.

Jameson Jensen · Accepted Answer

Step 1Assuming you still want to solve the problem using the method of Laplace transform. From your work, the Laplace transform Y(s) of y(t) satisfiesY(s)=1−e−ss(s+2)=1s(s+2)−e−ss(s+2).You can check that&amp;nbsp;1s(s+2)=12s−12(s+2).Step 2To transform the second term, you would need to use the so-called Translation on the t-axis property: If the inverse Laplace transform of F(s) is f(t), thenL−1{F(s)e−cs}=f(t−c)H(t−c),where&amp;nbsp;H(t−c)&amp;nbsp;is the shifted Heaviside function. You can check that y(t) is given byy(t)=12−12e−2t−H(t−1)[12−12e−2(t−1)].

Finding solution of ODE using Laplace transform with initial

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