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My Professor is asking us to solve the following differential equation and initial value problem but does not provide any further instructions. Could someone please help me get started?

1.) $D^3(D+5)(D+3)^2(D^2+16)y = 0$

and

2.)$2x e^y \mathrm{d}x+(x^2e^y+y \cos y)\mathrm{d}y = 0$

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  • $\begingroup$ Multiplication, i'll make the change $\endgroup$ – user100707 Dec 11 '13 at 16:07
  • $\begingroup$ Hint for $2.$, it is an Exact Equation. The first one could not me more straightforward. $\endgroup$ – Amzoti Dec 11 '13 at 16:08
  • $\begingroup$ Is solving a differential equation the same as providing a general solution for it? $\endgroup$ – user100707 Dec 11 '13 at 16:09
  • $\begingroup$ Please, stop writing latex with only $\$^y\$$, the dollar notation is meant to be used for a whole equation. Well, sorry, it's not especially after you, but it's a bad habit that seems to spread on MSE. $\endgroup$ – Jean-Claude Arbaut Dec 11 '13 at 16:10
  • $\begingroup$ @user100707: Yes, you can solve these, but do not have initial conditions, thus will have unknown constants floating about. $\endgroup$ – Amzoti Dec 11 '13 at 16:11
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According to The Annihilator Method, we have the following points for 1. :

  • When we have the term $D^3$ in the ODE so the terms $$c_1+c_2x+c_3x^2$$ will appear in $y(x)$.

  • When we have the term $D+5$ in the ODE so the terms $$c_4\exp(-5x)$$ will appear in $y(x)$.

  • When we have the term $(D+3)^2$ in the ODE so the terms $$c_5\exp(-3x)+c_6x\exp(-3x)$$ will appear in $y(x)$.

  • When we have the term $D^2+16$ in the ODE so the terms $$c_7\sin(4x)+c_8\cos(4x)$$ will appear in $y(x)$.

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1) If we set $$ P_1=D^3,\quad P_2=D+5,\quad P_3=(D+3)^2,\quad P_4=D^2+16, $$ then $P_iP_j=P_jP_i$ for every $i,j$. Therefore, the general solution $y$ of $$ D^3(D+5)(D+3)^2(D^2+16)y=P_1P_2P_3P_4y=0 $$ is given by $$ y=\mu_1\phi_1+\mu_2\phi_2+\mu_3\phi_3+\mu_4\phi_4, $$ where $\mu_1,\mu_2,\mu_3$ and $\mu_4$ are real constants and $$ P_1\phi_1=0,\quad P_2\phi_2=0,\quad P_3\phi_3=0,\quad P_4\phi_4=0. $$ It is obvious that \begin{eqnarray} \phi_1(x)&=&a_0+a_1x+a_1x^2,\quad \phi_2(x)=be^{-5x},\\ \phi_3(x)&=&(c_0+c_1x)e^{-3x},\quad \phi_4(x)=d_1\cos4x+d_2\sin4x. \end{eqnarray} Hence $$ y=C_0+C_1x+C_2x^2+C_3e^{-3x}+C_4e^{-5x}+C_5\cos4x+C_6\sin4x. $$ 2) Since $$ \frac{\partial}{\partial x}(x^2e^y+y\cos y)=2xe^y=\frac{\partial}{\partial y}(2xe^y), $$ it follows that there exists a function $(x,y) \mapsto f(x,y)$ such that $$ \frac{\partial f}{\partial x}(x,y)=2xe^y,\quad \frac{\partial f}{\partial y}(x,y)=x^2e^y+y\cos y. $$ From the identity $$ \frac{\partial f}{\partial x}(x,y)=2xe^y $$ we deduce that $$ f(x,y)=x^2e^y+g(y). $$ Since $$ \frac{\partial f}{\partial y}(x,y)=x^2e^y+g'(y)=x^2e^y+y\cos y, $$ it follows that $$ g'(y)=y\cos y. $$ Therefore $$ g(y)=\int y\cos y\,dy=\int y(\sin y)'\,dy=y\sin y-\int\sin y\,dy=y\sin y+\cos y+C. $$ Since $$ 2xe^ydx+(x^2e^y+y\cos y)dy=0 \iff df(x,y)=0 \iff f(x,y)=C', $$ where $C'$ is a real constant. It follows that the general solution of $$ 2xe^ydx+(x^2e^y+y\cos ydy)=0 $$ is the family of curves given by $$ x^2e^y+y\sin y+\cos y=c, $$ where $c$ is real constant.

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