Given that First solution is \(y_1=e^{2x}\)

let general solution \(y=v\cdot y_1\)

substitute \(y_1=e^{2x}\)

\(y=ve^{2x}\)

Differentiate \(y=ve^{2x}\) with respect to x

\(y'=\frac{d}{dx}(ve^{2x})\)

\(=\frac{vd}{dx}e^{2x}+e^{2x}\frac{d}{dx}(v)\)

\(=v\cdot2e^{2x}+e^{2x}\cdot v'\)

\(=2ve^{2x}+e^{2x}v'\)

hence, \(y'=2ve^{2x}+e^{2x}v,\)

Again differentiate \(y'=2ve^{2x}+e^{2x}v'\) with respect to x

\(y''=\frac{d}{dx(2ve^{2x}+e^{2x}v')}\)

\(=2\frac{d}{dx}(ve^{2x})+\frac{d}{dx}(e^{2x}v')\)

\(=4ve^{2x}+2e^{2x}v'+2e^{2x}v'+e^{2x}v''\)

\(=4ve^{2x}+4e^{2x}v'+e^{2x}v''\)

Hence, \(y''=4ve^{2x}+4e^{2x}v'+e^{2x}v''\)

Now, substitute \(y''=4ve^{2x}+4e^{2x}v'+e^{2x}v''\) and \(y=ve^{2x}\) in equation (1) and simlify it

\(4ve^{2x}+4e^{2x}v'+e^{2x}v''-4ve^{2x}=0\)

\(4e^{2x}v'+e^{2x}v''=0\) (2)

Let \(w=v'\)

\(\Rightarrow w'=v''\)

Substitute these value in equation (2)

\(4e^{2x}w+e^{2x}w'=0\)

\(e^{2x}(4w+w')=0\) (Take common as \(e^{2x}\))

Divide both side be \(e^{2x}\) and simplify it

\(\frac{e^{2x}(4w+w')}{e^{2x}}=\frac{0}{e^{2x}}\)

\(4w+w'=0\)

Substract 4w from both sides and simplify it

\(4w-4w+w'=-4w\)

\(w'=-4w\)

\(\frac{dw}{dx}=-4w\)

\({dw}{w}=-4dx\)

Now integrate it

\(\int\frac{dw}{w}=-4\int dx+c\)

\(\ln w=-4x+c\)

\(w=e^{-4x+c}\)

\(w=e^{-4x}\cdot e^c\)

Further simplify it

\(w=c_1e^{-4x}\) (Since \(e^c\) is constant say \(c_1\))

Now substitute \(w=v'\)

\(v'=c_1e^{-4x}\)

Take antiderivative

\(v=\frac{c_1e^{-4x}}{-4}+c_2\)

\(v=Ce^{-4}+c_2\) (Assume \(-\frac{c_1}{4}=C\))

Hence, \(v=Ce^{-4x}+c_2\)

Now substitute \(v=Ce^{-4x}+c_2\) in \(y=ve^{2x}\) and then multiply

\(y=(Ce^{-4x}+c_2)e^{2x}\)

\(y=Ce^{-4x}\cdot e^{2x}+c_2e^{2x}\)

\(y=Ce^{-4x+2x}+c_2e^{2x}\)

\(y=Ce^{-2x}+c_2e^{2x}\)

Hence, general solution is \(y=Ce^{-2x}+c_2e^{2x}\)

Therefore, second solution is \(y=e^{-2x}\)