# How to solve this differential equation using Laplace transform?

$$\dfrac{dy(t)}{dt}+\dfrac{1}{2}y(t).t=\dfrac{1}{2}$$ My attempt : $$sY(s)-\dfrac{1}{2}Y'(s)=\dfrac{1}{2s} \implies Y(s)=\dfrac{1}{2s^{2}}+\dfrac{1}{2s}Y'(s)$$ Now taking inverse laplace transform .. $$y(t)=\dfrac{1}{2}t.u(t)+\underbrace{\dfrac{1}{2} \displaystyle \int_{0}^{t}\left[ \displaystyle \int_{0}^{\infty} e^{st}.Y'(s)ds\right] \,dt}$$ This underbraced integral i'm not able to solve, and the result has erfi(x) on it . Please check my method and suggest something , thanks :)

• Do you know the rule for the laplace transform of the inverse? – user392395 Apr 7 '17 at 11:25
• yeah i know @Abdel – zeno-san Apr 7 '17 at 11:26
• consider this L{f′(t)}=sY(s)−f(0) – user392395 Apr 7 '17 at 11:27
• yeah i know this $$\dfrac{dy(t)}{dt} \rightarrow sY(s)$$ no initial conditions – zeno-san Apr 7 '17 at 11:28
• you should thake f(0) equal to a constant C – user392395 Apr 7 '17 at 11:36