# One solution of the differential equation y" – 4y = 0 is y = e^{2x} Find a second linearly independent solution using reduction of order.

Question
One solution of the differential equation $$y" – 4y = 0$$ is $$y = e^{2x} Find a second linearly independent solution using reduction of order. ## Answers (1) 2021-01-29 Given differential equation is \(y''-4y=0$$ (1)
Given that First solution is $$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}$$

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