Question

The differential equation

dy/dx= cos(x)(y^2 +6y+8)/(6y+16)

has an implicit general solution of the form F(x,y)=K

In fact, because the differential equation is separable, we can define the solution curve implicitly by a function in the form

F(x,y)=G(x)+H(y)=K

Find such a solution and then give the related functions requested.

F(x,y)=G(x)+H(y)=

Answer 1

The given equation can be separated as, $\cos x~ dx = \dfrac{6y+16}{(y+2)(y+4)}dy$.

Now using partial fractions for the term on the right hand side,

$\begin{aligned} \frac{6y+16}{(y+2)(y+4)} & = \frac{A}{y+2} + \frac{B}{y+4}\\ 6y+16 & =A(y+4) + B(y + 2)\\ \end{aligned}$

Taking $y = -4$ gives, $B = 4$ and taking $y = -2$ gives, $A = 2$.

Therefore,

$\cos x~ dx = \left\{\dfrac{2}{y+2} + \dfrac{4}{y+4}\right\}dy$

Integrating both sides, we get

$\begin{aligned} \sin x &= 2\log (y+2) + 4\log (y+4) \\ \sin x &= \log (y+2)^2 + \log (y+4)^4\\ \sin x + c &= \log \{(y+2)^2 (y+4)^4\}\\ e^{(\sin x) + c} &= (y+2)^2(y+4)^4\\ (y+2)^2(y+4)^4 - e^{\sin x} &= c_1~~~~(\text{where }c_1 = e^c) \end{aligned}$

Hence, $F(x,y)=G(x)+H(y)= - e^{\sin x} +(y+2)^2(y+4)^4$ where

$G(x)= - e^{\sin x}; H(y) = (y+2)^2(y+4)^4$.