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$$\dfrac{dy(t)}{dt}+\dfrac{1}{2}y(t).t=\dfrac{1}{2} $$ 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 :)

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

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