Question

Please solve problem below​

Answer 1

(a) This is a Bernoulli equation:

$\dfrac{dy}{dx} + \dfrac{x}{2(x^2+1)} y = x^3y^5$

$\implies y^{-5} \dfrac{dy}{dx} + \dfrac{x}{2(x^2+1)} y^{-4} = x^3$

Substitute $z=y^{-4}$ and $\frac{dz}{dx} = -4y^{-5}\frac{dy}{dx}$ to transform the ODE to

$-\dfrac14 \dfrac{dz}{dx} + \dfrac{x}{2(x^2+1)} z = x^3$

$\implies \dfrac{dz}{dx} - \dfrac{2x}{x^2+1} z = -4x^3$

which is now linear in $z$. Using the integrating factor method, the I.F. is

$\mu = \displaystyle \exp\left(\int -\frac{x}{x^2+1} \, dx\right) = \exp\left(-\frac12 \ln(1+x^2)\right) = \dfrac1{\sqrt{1+x^2}}$

Distribute $\mu$ on both sides to get a derivative of a product on the left side.

$\dfrac1{\sqrt{1+x^2}} \dfrac{dz}{dx} - \dfrac{2x}{(x^2+1)^{3/2}} z = -\dfrac{4x^3}{\sqrt{1+x^2}}$

$\implies \dfrac{d}{dx} \left(\dfrac1{\sqrt{1+x^2}} z\right) = -\dfrac{4x^3}{\sqrt{1+x^2}}$

Integrate both sides (the integral on the right can be done by parts) to get

$\displaystyle \frac1{\sqrt{1+x^2}} z = -4 \int \frac{x^3}{\sqrt{1+x^2}} \, dx = -\frac43 (x^2-2) \sqrt{1+x^2} + C$

Solve for $z$.

$\displaystyle \frac1{\sqrt{1+x^2}} z = -\frac43 (x^2-2) \sqrt{1+x^2} + C$

$\implies z = -\dfrac43 (x^2-2) (1+x^2) + C \sqrt{1+x^2}$

Solve for $y$.

$\dfrac1{y^4} = -\dfrac43 (x^2-2) (1+x^2) + C \sqrt{1+x^2}$

$\implies \boxed{y^4 = -\dfrac3{4(x^2-2) (1+x^2) + C \sqrt{1+x^2}}}$

You could go on to solve explicitly for $y$ if you like.

(b) This is also a Bernoulli equation:

$x^2y' + 2xy - y^3 = 0$

$\implies x^2 y^{-3} y' + 2xy^{-2} = 1$

Substitute $z=y^{-2}$ and $z' = -2y^{-3}y'$.

$-\dfrac{x^2}2 z' + 2xz = 1$

$\implies z' - \dfrac4x z = -\dfrac2{x^2}$

Now repeat the method from (a) to solve for $y$.

$\mu = \exp\left(-\displaystyle \int \frac4x \, dx\right) = \dfrac1{x^4}$

$\implies \dfrac1{x^4} z' - \dfrac4{x^5} z = -\dfrac2{x^6}$

$\implies \left(\dfrac1{x^4} z\right)' = -\dfrac2{x^6}$

$\displaystyle \implies \dfrac1{x^4} z = -2 \int \frac{dx}{x^6}$

$\implies \dfrac1{x^4} z = \dfrac2{5x^5} + C$

$\implies z = \dfrac2{5x} + Cx^4$

$\implies \dfrac1{y^2} = \dfrac2{5x} + Cx^4$

$\implies \boxed{y^2 = \dfrac{5x}{2 + Cx^5}}$