Question

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

Second order linear equations
ANSWERED
asked 2021-01-28

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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