\frac{dy}{dt} - \frac{1}{2}y = e^\frac{1}{2}t \sin t

Question:

Solution to Differential Equation:

Differential equation with form {eq}\displaystyle \frac{dy}{dt}+Py=Q {/eq} is called linear differential equation.

Here, P and Q are constants or function of t only.

It's solution will be,

{eq}\displaystyle \:y\left(I.F\right)=\int \:Q\left(I.F\right)dt+C {/eq} with {eq}\displaystyle I.F=e^{\int \:P\:dt} {/eq}

Answer and Explanation: