1 problem found
Let \(P,Q\) and \(R\) be functions of \(x\). Prove that, for any function \(y\) of \(x\), the function \[ Py''+Qy'+Ry \] can be written in the form \(\dfrac{\mathrm{d}}{\mathrm{d}x}(py'+qy),\) where \(p\) and \(q\) are functions of \(x\), if and only if \(P''-Q'+R=0.\) Solve the differential equation \[ (x-x^{4})y''+(1-7x^{3})y'-9x^{2}y=(x^{3}+3x^2)\mathrm{e}^{x}, \] given that when \(x=2,y=2\mathrm{e}^{2}\) and \(y'=0.\)
Solution: Suppose \(Py'' + Qy' + Ry = \frac{\d}{\d x}(p y' + qy)\), then \begin{align*} Py'' + Qy' + Ry &= \frac{\d}{\d x}(p y' + qy) \\ &= py'' + p'y' + qy' + q' y \\ &= py'' + (p'+q)y' + q' y \end{align*} Therefore \(P = p, Q = p'+q, R = q'\), Therefore \(q = Q-P'\) and \(R = Q'-P''\) or \(P'' -Q'+R = 0\). \((\Rightarrow)\) Suppose it can be written in that form, then the logic we have applied shows that equation is true. \((\Leftarrow)\) Suppose \(P''-Q'+R = 0\), then letting \(p = P, q = Q-P'\) we find functions of the form which will be expressed correctly. Notice that if \(P = x-x^4, Q = (1-7x^3), R = -9x^2\) then: \begin{align*} P'' - Q' + R &= -12x^2+21x^2-9x^2 \\ &= 0 \end{align*} Therefore we can write our second order ODE as: \begin{align*} && (x^{3}+3x)\mathrm{e}^{x} &= \frac{\d}{\d x} \left ( (x-x^4) y' +(1-7x^3-(1-4x^3))y \right) \\ &&&= \frac{\d}{\d x} \left ( (x-x^4) y' -3x^3y \right) \end{align*} Suppose \(z = (x-x^4)y' - 3x^2y\), then \(z = (2-2^4) \cdot 0 - 3 \cdot 2^2 \cdot 2e^2 = -24e^2\) when \(x = 2\). and we have: \begin{align*} && \frac{\d z}{\d x} &= (x^3+3x^2)e^x \\ \Rightarrow && z &= \int (x^3+3x^2)e^x \d x \\ &&&= x^3 e^x+c \\ \Rightarrow && -48e^2 &= e^2(8) + c \\ \Rightarrow && c &= -56e^2 \\ \Rightarrow && z &= e^x(x^3)-56e^2 \\ \end{align*} So our differential equation is: \begin{align*} && (x-x^4)y'-3x^3 y &= x^3e^x -5 6e^2 \\ \Rightarrow && (1-x^3)y' -3x^2y &= x^2 e^x - \frac{6e^2}{x} \\ \Rightarrow && \frac{\d }{\d x} \left ( (1-x^3)y \right) &= x^2e^x - \frac{56e^2}{x} \\ \Rightarrow && (1-x^3)y &= (x^2-2x+2)e^x - 56e^2 \ln x + k \\ \underbrace{\Rightarrow}_{x=2} && (1-2^3)2e^2 &= (2^2-2\cdot2 + 2)e^2 -56e^2 \ln 2 + k \\ \Rightarrow && k &= -16e^2+56 \ln 2 \cdot e^2 \\ \Rightarrow && y &= \frac{(x^2-2x+2)e^x - 56e^2 \ln x -16e^2+56 \ln 2 \cdot e^2}{(1-x^3)} \end{align*}