Solve this differential equation $dx/dt=x^2t^3+xt$

This is not exact.I don't find any way to solve this by integrating factor

  • 1
    $\begingroup$ This is a Bernoulli equation. The standard method is to substitute $z = x^{-1}$ $\endgroup$ – Dylan Jan 25 at 14:47

$$\frac{dx}{dt}=x^2t^3+xt$$ $$(x^2t^3+xt)dt-dx=0$$ Solving thanks to the integrating factor method.

Let $\mu(x,t)$ an integrating factor. $$(x^2t^3+xt)\mu dt-\mu dx=0$$ The condition to be an exact differential is : $$\frac{\partial}{\partial x}\left((x^2t^3+xt)\mu \right) = \frac{\partial}{\partial t}(-\mu)$$ $$\frac{\partial \mu}{\partial t}+(x^2t^3+xt)\frac{\partial \mu}{\partial x}+(2xt^3+t)\mu=0$$ In the most simple cases $\mu(x,t)$ is on the form $f(x)$ or $g(t)$ or $h(xt)$. But in the present case, trying those forms of $\mu(x,y)$ fails. So, we try a more general form, for example $\mu(x,t)=f(x)g(t)$ $$f(x)\frac{dg}{dt}+(x^2t^3+xt)g(t)\frac{df}{dx}+(2xt^3+t)f(x)g(t)=0$$ $$\frac{1}{tg(t)}\frac{dg}{dt}+(xt^2+1)\frac{x}{f(x)}\frac{df}{dx}+(2xt^2+1)=0$$ The separation of variables becomes possible if $$xt^2\frac{x}{f(x)}\frac{df}{dx}+2xt^2=0\quad\implies\quad \frac{x}{f(x)}\frac{df}{dx}=-2\quad\implies\quad f(x)=\frac{1}{x^2}$$ Then $$\frac{1}{tg(t)}\frac{dg}{dt}+\frac{x}{f(x)}\frac{df}{dx}+1=0 \quad;\quad \frac{1}{tg(t)}\frac{dg}{dt}-2+1=0$$ $$\frac{1}{g(t)}\frac{dg}{dt}=t\quad\implies\quad g(t)=e^{t^2/2}$$ An integrator factor is : $$\mu(x,t)=\frac{e^{t^2/2}}{x^2}$$ From this it is easy to solve the ODE. The result is : $$e^{t^2/2}\left(\frac{1}{x}+t^2-2 \right)=c$$ $$x(t)=\frac{-1}{2-t^2+c\:e^{-t^2/2}}$$


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