Answer to Question #163343 in Differential Equations for ali hassana

Question

Solve and find particular solution of the first order non-linear differential equation of Bernoulli’s type with given condition.

(dy/dx)+(xy/1-x² )=xy^1/2 ; y(0)=1

 

Answer 1

Solve and find particular solution of the first order non-linear differential equation of Bernoulli’s type with given condition.

(dy/dx)+(xy/1-x² )=xy^1/2 ; y(0)=1

Solution:

Divide the left and right sides of the equation by $\sqrt y$ :

$\displaystyle\frac{1}{\sqrt y}\frac{dy}{dx}+\frac{x}{1-x^2}\sqrt y=x$

$\sqrt{y}=z$ , $z'=\displaystyle\frac{y'}{2 \sqrt y}$ , $\displaystyle\frac{y'}{\sqrt y}=2z'$ .

$\displaystyle2z'+\frac{x}{1-x^2}z=x$

$z=uv$ , $z'=u'v+v'u$

$\displaystyle2(u'v+v'u)+\frac{x}{1-x^2}uv=x$

$\displaystyle2u'v+u(2v'+\frac{x}{1-x^2}v)=x$

Let's compose and solve the system:

$\begin{cases} 2v'+\frac{x}{1-x^2}v=0 \\ 2u'v=x \end{cases}$

From the first equation:

$2v'=-\frac{x}{1-x^2}v$

$2\frac{dv}{v}=-\frac{xdx}{1-x^2}$

$2\frac{dv}{v}=-\frac12\frac{dx^2}{1-x^2}$

$4\ln{|v|}=\ln{|1-x^2|}$

$v=(1-x^2)^{\frac14}$

Substitute $v$ into the second equation:

$2u'(1-x^2)^{\frac14}=x$

$u=\displaystyle\int\frac{xdx}{2(1-x^2)^{\frac14}}=\displaystyle\int\frac{dx^2}{4(1-x^2)^{\frac14}}=-\frac13(1-x^2)^{\frac34}+C$

$z=uv=(1-x^2)^{\frac14}(-\frac13(1-x^2)^{\frac34}+C)=-\frac13(1-x^2)+C(1-x^2)^{\frac14}$

$\sqrt y=-\frac13(1-x^2)+C(1-x^2)^{\frac14}$

$y(0)=1$

$\sqrt 1=-\frac13(1-0^2)+C(1-0^2)^{\frac14}$

$1=-\frac13+C$

$C=\frac43$

$\sqrt y=-\frac13(1-x^2)+\frac43(1-x^2)^{\frac14}$

Answer: $\sqrt y=-\frac13(1-x^2)+\frac43(1-x^2)^{\frac14}$ .