First order non linear ODE with Bernoulli

Question

I have a problem with this equation: $ y'(x)-xy(x)=-xy^4(x) $ with initial condition $ y(x_{0})=y_{0}$. I'm arrived to prove that $ y_{0}= (Ce^{-\frac{3}{2}x_{0}^{2}}+1)^{-3} $ but now i can't move on. Moreover, WolframAlpha's solution is next to impossible…

Thanks for any help!

Answer 1

Hint: Write this as $$\frac{y'(x)}{y(x)-y(x)^4}=x$$

Answer 2

We have $$C=e^{\frac{3{x_0}^2}{2}}({y_0}^{-1/3}-1)$$

Answer 3

Let us solve the Bernoulli differential equation $$y'(x)-xy(x)=-xy(x)^4$$ with initial condition $y (x_0)=y_0$. Dividing by $y^4$ and setting $u=y^{-3}$, we have the linear ODE $$ \frac{1}{3} u'(x)+xu(x) = x $$ with initial condition $u (x_0)={y_0}^{-3}$. The solution obtained by integrating factor reads \begin{aligned} u(x) &= e^{-3 (x^2-{x_0}^2)/2} \left({y_0}^{-3} + 3 \int_{x_0}^x t e^{3t^2/2} \,\text d t \right) \\ &= e^{-3 (x^2-{x_0}^2)/2} \left({y_0}^{-3} + e^{3x^2/2} - e^{3{x_0}^2/2} \right) , \end{aligned} from which one deduces $y=u^{-1/3}$.

Answer 4

Substitute $z=1/y^3$ $$z'+3xz=3x$$ Tis equation is separable $$z'=3x(1-z)$$ $$\int \frac {dz}{1-z}=\frac 32{x^2}+C$$ $$-\ln ({z-1})=\frac 32{x^2}+C$$ $$\implies y^3(x)=\frac 1 {Ke^{-3x^2/2}+1}$$

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Therefore $$y_0^3=\frac 1 {Ke^{-3x_0^2/2}+1}$$ $${Ke^{-3x_0^2/2}}=\frac 1 {y_0^3}-1$$ $$K=\left (\frac 1 {y_0^3}-1 \right)e^{3x_0^2/2}$$

Answer 5

To reach a better understanding of my problem, i will write all the passages.

$\frac{y^{'}}{y^{4}}=\frac{xy}{y^{3}}-x\rightarrow \frac{y^{'}}{y^{4}}=xy^{-3}-x$

Now i put $z=y^{-3}\rightarrow z^{'}=-3xz+3x$.

$y_{0}(x)=Ce^{A(x)}\rightarrow A(x)=\int -3xdx=-\frac{3}{2}x^{2}\rightarrow y_{0}(x)=Ce^{-\frac{3}{2}x^{2}}$

$y_{p}(x)=e^{A(x)}B(x)\rightarrow B(x)=\int -3x\cdot e^{A(x)}dx=\int -3x\cdot e^{\frac{3}{2}x^{2}}dx$

Now i put $\frac{3}{2}x^{2}=t\rightarrow dt=3xdx\rightarrow dx=\frac{dx}{3x}$

Since $\int -3xe^{t}\frac{dt}{3x}=-e^{\frac{3}{2}x^{2}}\rightarrow y_p{x}=1$, i obtain $y(x)=Ce^{-\frac{3}{2}x^{2}}+1$. But since $y^{-3}=z$...the result that i wrote.