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

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)

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 \(y_1=e^{2x}\)
Differentiate \(y=ve^{2x}\) with respect to x
\(=v\cdot2e^{2x}+e^{2x}\cdot v'\)
hence, \(y'=2ve^{2x}+e^{2x}v,\)
Again differentiate \(y'=2ve^{2x}+e^{2x}v'\) with respect to x
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
\(4e^{2x}v'+e^{2x}v''=0\) (2)
Let \(w=v'\)
\(\Rightarrow w'=v''\)
Substitute these value in equation (2)
\(e^{2x}(4w+w')=0\) (Take common as \(e^{2x}\))
Divide both side be \(e^{2x}\) and simplify it
Substract 4w from both sides and simplify it
Now integrate it
\(\int\frac{dw}{w}=-4\int dx+c\)
\(\ln w=-4x+c\)
\(w=e^{-4x}\cdot e^c\)
Further simplify it
\(w=c_1e^{-4x}\) (Since \(e^c\) is constant say \(c_1\))
Now substitute \(w=v'\)
Take antiderivative
\(v=Ce^{-4}+c_2\) (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
\(y=Ce^{-4x}\cdot e^{2x}+c_2e^{2x}\)
Hence, general solution is \(y=Ce^{-2x}+c_2e^{2x}\)
Therefore, second solution is \(y=e^{-2x}\)

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