I have a question where I need to find the general solution of the differential equation $y-3y^{7}=(y^{3}+6x)y'$, where the solution is in the form $F(x,y)=C$. I am only concerned about finding $F(x,y)$, and I have been advised to rewrite this equation in differential form, but I am still unsure where to go from here. Any help would be greatly appreciated with this problem!
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$\begingroup$ you must compute a integrating factor $\endgroup$– Dr. Sonnhard GraubnerCommented Oct 1, 2017 at 16:05
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3 Answers
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Hint:
Write it in terms of $x'(y)$. You will see its a linear differential equation!
$$\frac{dx}{dy} + \frac{6x}{y-3y^7} = \frac{y^3}{y-3y^7}$$
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$$\frac{d\mu(y)}{dy}(-3y^7+y)(-21y^6+1)\mu(y)=-6\mu(y)$$ $$\frac{\frac{\partial \mu(y)}{\partial y}}{\mu(y)}=-\frac{7}{y}$$ $$\mu(y)=\frac{1}{y^7}$$ and then we have $$-3+\frac{1}{y(x)^6}-\frac{(6x+y(x)^3)y'(x)}{y(x)^7}=0$$ and let $$P(x,y)=\frac{1}{y^6}-3,Q(x,y)=-\frac{6x+y^3}{y^7}$$ and we get $$\frac{\partial P(x,y}{\partial y}=-\frac{6}{y^7}=\frac{\partial Q(x,y)}{\partial x}$$ and the solution is given by $$f(x,y)=C_1$$ where $C_1$ is a arbitrary constant
answered Oct 1, 2017 at 16:08
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$\begingroup$ Would I then just multiply $y-3y^{7}=(y^{3}+6x)y'$ with the integrating factor $\frac{1}{y^{7}}$ ? $\endgroup$ Commented Oct 1, 2017 at 16:19
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$\begingroup$ yes this would i do $\endgroup$ Commented Oct 1, 2017 at 16:21
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$\begingroup$ So in this case, once we apply the integrating factor, we can then treat it like an exact differential? $\endgroup$ Commented Oct 1, 2017 at 16:44
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$\begingroup$ @Dr Can you please explain to me your method of solution! $\endgroup$– jonsnoCommented Oct 1, 2017 at 17:29
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Hint:
Write $(y-3y^7)dx-(y^3+6x)dy=0$ then $$p(y)=\dfrac{M_y-N_x}{-M}=\dfrac{7}{-y}$$ then $$I=e^{\int p(y)dy}=\dfrac{1}{y^7}$$ is integrating factor.
