Question #105504
a) Solve the differential equation
dy/dx+(x/1-x^2)y=x (sqrt y),y(0)=1
1
Expert's answer
2020-03-27T14:32:08-0400

Given Differential equation is,


dydx+(x1x2)y=x×y\frac{dy}{dx} +( \frac {x}{1 -x^2}) y= x \times \sqrt y

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



1ydydx+(x1x2)y=x.............(A)\frac {1}{\sqrt y} \frac{dy}{dx} + (\frac {x}{1 -x^2}) \sqrt y = x .............(A)

Let y=t\sqrt y = t

Differentiate with respect to xx

Then,

12ydydx=dtdx\frac {1}{2\sqrt y} \frac {dy}{dx} = \frac {dt}{dx}

Then, the equation (A) becomes



2dtdx+(x1x2)t=x2 \frac {dt}{dx} + (\frac {x}{1 -x^2}) t = x

divide by 2


dtdx+12(x1x2)t=x2\frac {dt}{dx} + \frac {1}{2} (\frac {x}{1 -x^2}) t = \frac {x}{2}

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


Integrating factor = I.F = ep(x)dx=e12(x1x2)dxe^ {\int p(x) dx} = e^ {\int \frac {1}{2} (\frac {x}{1 -x^2})dx}


=e142x1x2dx=e14ln(1x2)=11x24= 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×(I.F)=(I.F)×Q×dx+ct \times (I.F)= \int (I.F) \times Q \times dx + c

t×11x24=11x24×x2 dx+ct \times \frac {1}{\sqrt[4] {1- x^2}} = \int \frac {1}{\sqrt[4] {1- x^2}} \times \frac {x}{2} \space dx +c

 



t1x24=142x1x24dx+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 1x2=m then 2xdx=dm1 - x^2 = m \space then \space - 2x dx = dm


t1x24=141m4dm+c\frac {t}{\sqrt [4] {1-x^2}} = -\frac {1}{4}\int \frac {1}{\sqrt [4]{m}} dm + c

t1x24=14×m14+114+1+c\frac {t}{\sqrt [4] {1-x^2}} = -\frac {1}{4} \times \frac {m^ {\frac {1}{4} + 1}}{\frac {1}{4}+1}+ c

t1x24=44×m545+c\frac {t}{\sqrt [4] {1-x^2}} = -\frac {4}{4} \times \frac {m^ {\frac {5}{4} }}{5}+ c

t1x24=(1x2)545+c\frac {t}{\sqrt [4] {1-x^2}} = -\frac {(1-x^2)^ {\frac {5}{4} }}{5}+ c

y(0)=1y(0)=1



1104=(10)545+c\frac {1}{\sqrt [4] {1-0}} = -\frac {(1-0)^ {\frac {5}{4} }}{5}+ c

1=15+cc=1+15=651 = \frac{-1}{5} + c \\ c = 1 + \frac {1}{5} = \frac {6}{5}

So the solution of the equation is,


y1x24=(1x2)545+65\frac {\sqrt {y}}{\sqrt [4] {1-x^2}} = -\frac {(1-x^2)^ {\frac {5}{4} }}{5}+ \frac{6}{5}



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Comments

Assignment Expert
16.04.20, 17:04

Dear Himanshu, You are welcome. We are glad to be helpful. If you liked our service, please press a like-button beside the answer field. Thank you!

Himanshu
16.04.20, 11:58

Thanks a lot for help.

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