Answer to Question #105504 in Differential Equations for khushi

Question

a) Solve the differential equation
dy/dx+(x/1-x^2)y=x (sqrt y),y(0)=1

Answer 1

Given Differential equation is,

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

Divide by $\sqrt y$ on both sides of the above equation, then we get

$$\frac {1}{\sqrt y} \frac{dy}{dx} + (\frac {x}{1 -x^2}) \sqrt y = x .............(A)$$

Let $\sqrt y = t$

Differentiate with respect to $x$

Then,

$$\frac {1}{2\sqrt y} \frac {dy}{dx} = \frac {dt}{dx}$$

Then, the equation (A) becomes

$$2 \frac {dt}{dx} + (\frac {x}{1 -x^2}) t = x$$

divide by 2

$$\frac {dt}{dx} + \frac {1}{2} (\frac {x}{1 -x^2}) t = \frac {x}{2}$$

This is the Linear Differential equation in the form $\frac {dt}{dx} + P(x) t= Q(x)$

Integrating factor = I.F = $e^ {\int p(x) dx} = e^ {\int \frac {1}{2} (\frac {x}{1 -x^2})dx}$

$$= e^{\frac {1}{4}\int \frac {2x}{1-x^2}dx } = e^ {- \frac {1}{4} ln {( 1- x^2) }} = \frac {1}{\sqrt[4] {1-x^2}}$$

Now the solution of the linear differential equation is

$t \times (I.F)= \int (I.F) \times Q \times dx + c$

$$t \times \frac {1}{\sqrt[4] {1- x^2}} = \int \frac {1}{\sqrt[4] {1- x^2}} \times \frac {x}{2} \space dx +c$$

 

$$\frac {t}{\sqrt [4] {1-x^2}} = -\frac {1}{4}\int \frac {-2x}{\sqrt [4]{1-x^2}} dx + c$$

use substitution method on right side Integration,

Let $1 - x^2 = m \space then \space - 2x dx = dm$

$$\frac {t}{\sqrt [4] {1-x^2}} = -\frac {1}{4}\int \frac {1}{\sqrt [4]{m}} dm + c$$

$$\frac {t}{\sqrt [4] {1-x^2}} = -\frac {1}{4} \times \frac {m^ {\frac {1}{4} + 1}}{\frac {1}{4}+1}+ c$$

$$\frac {t}{\sqrt [4] {1-x^2}} = -\frac {4}{4} \times \frac {m^ {\frac {5}{4} }}{5}+ c$$

$$\frac {t}{\sqrt [4] {1-x^2}} = -\frac {(1-x^2)^ {\frac {5}{4} }}{5}+ c$$

$y(0)=1$

$$\frac {1}{\sqrt [4] {1-0}} = -\frac {(1-0)^ {\frac {5}{4} }}{5}+ c$$

$1 = \frac{-1}{5} + c \\ c = 1 + \frac {1}{5} = \frac {6}{5}$

So the solution of the equation is,

$$\frac {\sqrt {y}}{\sqrt [4] {1-x^2}} = -\frac {(1-x^2)^ {\frac {5}{4} }}{5}+ \frac{6}{5}$$