Method of variation of parameter

Question

By method of variation of parameters find the PI of the differential equation $$\frac{dy}{dx}+2y=4e^{2x}$$ I have learned how to do the find out the PI of second order differential equation. But, here is first order. I can't find any solution of this. Please help me. $$\text{I tried to use the Wronskian method like:}$$ $$Let, y=e^{mx}\\ \implies Dy=me^{mx}$$ $$\text{Then, the auxiliary equation is:}\\ m+2=0\\ \implies m=-2$$ $$Then, C.F.=c_1e^{mx}$$ $$P.I. y_p= u(x)e^{mx}$$ Then, there we had to find wronskian. But, I can't get it how to find that...

Answer 1

You don't need the Wronskian here it's a first order DE $$\frac{dy}{dx}+2y=4e^{2x}$$ Solve the homogeneous equation $$\frac{dy}{dx}+2y=0$$ $$\ln y =-2\int dx=-2x+c$$ $$y=ce^{-2x}$$ Variation of parameter needs $$y=c(x)e^{-2x}$$ $$y'=c'(x)e^{-2x}-2e^{-2x}c(x)$$ Plug this in the differential equation and solve for $c(x)$. $$y'+2y=4e^{2x}$$ $$c'(x)e^{-2x}-2e^{-2x}c(x)+2c(x)e^{-2x}=4e^{2x}$$ Can you take it from here ?

Answer 2

Consider a general scalar dynamical system where $a,b \in \mathbb{R}$ $$\dot{x}(t)=ax(t)+bu(t) \implies \dot{x}(t)-ax(t)=u(t)$$ Now multiply both sides by $e^{-at}$, you get $$e^{-at} \left( \dot{x}(t)-ax(t)\right)=\frac{\mathrm{d}}{\mathrm{d}t}e^{-at}x(t)=e^{-at}bu(t)$$Now integrating this between $0$ to $t$ gives you $$x(t)=e^{at}x(0)+e^{at}\int_0^t e^{-as}b u(s)\mathrm{d}s$$ Where the first term is the response due to initial condition $x(0)$ and the second term is the forced response due to the input $u(t)$