1 problem found
Let \(y,u,v,P\) and \(Q\) all be functions of \(x\). Show that the substitution \(y=uv\) in the differential equation \[ \frac{\mathrm{d}y}{\mathrm{d}x}+Py=Q \] leads to an equation for \(\dfrac{\mathrm{d}v}{\mathrm{d}x}\) in terms of \(x,Q\) and \(u\), provided that \(u\) satisfies a suitable first order differential equation. Hence or otherwise solve \[ \frac{\mathrm{d}y}{\mathrm{d}x}-\frac{2y}{x+1}=\left(x+1\right)^{\frac{5}{2}}, \] given that \(y(1)=0\). For what set of values of \(x\) is the solution valid?
Solution: Suppose \(y = uv\) then and suppose \(\frac{\d u}{\d x} + P u = 0\) then \begin{align*} && \frac{\d y}{\d x} + Py &= Q \\ && uv' + u'v + Puv &= Q \\ && uv' &= Q \\ && \frac{\d v}{\d x} &= \frac{Q}{u} \end{align*} Consider \begin{align*} && 0 &= \frac{\d u}{\d x} - \frac{2u}{x+1} \\ \Rightarrow && \ln u &= 2\ln (1 + x) + C \\ \Rightarrow && u &= A(1+x)^2 \end{align*} and \begin{align*} && \frac{\d v}{\d x} &= \frac1{A}(x+1)^{\frac12} \\ \Rightarrow && v &= \frac2{3A}(x+1)^{\frac32} + k \\ \Rightarrow && y &= \frac23(x+1)^\frac72 + k(x+1)^2 \\ && 0 &= y(1) \\ &&&= \frac23 2^{7/2}+k2^2 \\ \Rightarrow && k &= -\frac{2^{5/2}}{3} \\ \Rightarrow && y &= \frac23 (x+1)^{7/2} - \frac{2^{5/2}}{3}(x+1)^2 \end{align*}