Exercise · ZW-VARIATION-OF-PARAMETERS-009 · ★★★ Advanced
With \(y_1=e^{t}\), \(y_2=te^{t}\) (\(W=e^{2t}\)), find \(y_p\) for \(y''-2y'+y=\dfrac{e^{t}}{t}\), \(t>0\), taking \(u_1\) and \(u_2\) with no constant.
Choose one option.
A \(te^{t}\ln t-te^{t}\) B \(te^{t}\ln t-t^2e^{t}\) C \(e^{t}\ln t-te^{t}\) D \(t^2e^{t}\ln t\) Show answer A. \(te^{t}\ln t-te^{t}\)← Back to the question
Explanation \(u_1'=-\frac{y_2g}{W}=-1\) and \(u_2'=\frac{y_1g}{W}=\frac1t\), so \(u_1=-t\), \(u_2=\ln t\). Then \(y_p=-te^t+te^t\ln t\).
A method that builds a particular solution \(y_p\) of \(y''+py'+qy=g(t)\) from two homogeneous solutions \(y_1,y_2\) and their Wronskian \(W\), by letting the constants vary.
\[y_p=-y_1\!\int\!\frac{y_2\,g}{W}\,dt+y_2\!\int\!\frac{y_1\,g}{W}\,dt\]
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