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The Wronskian of \(y_1=x\) and \(y_2=e^x\) is \(W=y_1y_2^{\,\prime }-y_1^{\prime }y_2=(x-1)e^x\). Then there is a particular solution to \((x-1)y^{\,\prime \prime }-xy^{\,\prime }+y=2(x-1)^2\,e^{x}\) of the
form \(y_p=uy_1+vy_2\) where \(\displaystyle u^{\,\prime }=-\frac {2(x-1)^2e^{x}e^x}{(x-1)e^x}=-2(x-1)e^{x}\) and
\(\displaystyle v^{\,\prime }=\frac {2(x-1)^2e^{x}x}{(x-1)e^x}=2x^2-2x\). By integrating, we get \(u=-2(x-2)e^x\) and \(\displaystyle v=\frac {2}{3}x^3-x^2\). So a solution is
\(y_p=ux+ve^x=(-2(x-2)e^x)x+\left (\frac {2}{3}x^3-x^2\right )e^x=\left (\frac {2}{3}x^3-3x^2+4x\right )e^x\).
Hint: you may find \(\displaystyle \int \sec (\theta )\, \mathrm {d}\theta =\ln \left |\sec (\theta )+\tan (\theta )\right |+C\) helpful.