Answer to Question #116836 in Differential Equations for roopak

Question #116836
solve the differential equation dy/dx + (x/1-x^2)y=xy^1/2 , y(0)=1

correct question
1
Expert's answer
2020-05-19T08:56:48-0400

the given equation can be written as

1ydydx+2xy2(1x2)=x\frac{1}{\sqrt{y}} \frac{dy}{dx} + \frac{2x\sqrt{y}}{2(1-x^2)} = x .....(1)


this is of the form f'(y) y' + P f(y) = Q

where P = x2(1x2)\frac{x}{2(1-x^2)}

Q= x

f(y) = 2 y\sqrt{y}


let v= 2y\sqrt{y}

differentiating,

1ydydx=dvdx\frac{1}{\sqrt{y}} \frac{dy}{dx} = \frac{dv}{dx}


(1) becomes

dvdx+xv2(1x2)=x\frac{dv}{dx} + \frac{xv}{2(1-x^2)} = x


this is of the form dvdx+Av=B\frac{dv}{dx} + Av = B


A=x2(1x2)andB=xA = \frac{x}{2(1-x^2)} and B=x


integrating factor = ePdx=ex2(1x2)dx=(1x2)(1/4)e^{\intop Pdx} = e^{\intop \frac{x}{2(1-x^2)}dx} = (1-x^2)^{(-1/4)}


the general solution is given by:

v (1-x2)-1/4 = \int x. (1-x2)-1/4 dx + c


2y\sqrt{y} = -2/3 (1-x2) + c(1-x2)1/4

now, given y(0)=1


2 = -2/3 + c

c= 8/3


so, the solution is

2y\sqrt{y} = -2/3 (1-x2) + 8/3(1-x2)1/4

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment