Answer to Question #177757 in Differential Equations for Zain Ul Abideen Khan

Question #177757

Solve by finding an integrating factor

(x^2+y^2+2x)dx + 2ydy =0


1
Expert's answer
2021-04-15T07:20:54-0400

Solution


given the differential equation:

(x2+y2+2x)dx+2ydy=0(x^2+y^2+2x)dx+2ydy=0 ...(1)

let f(x,y)=x2+y2+2xf(x,y)=x^2+y^2+2x

    g(y)=2y\implies g(y)=2y


so

f'y(x,y)=2y_y(x,y)=2y

g(x)=0g'(x)=0


using Integration factor method,

dudx=(fygx)ug\frac{du}{dx}=\frac{(f_y-g_x)u}{g}

dudx=(2y0)u2y\frac{du}{dx}=\frac{(2y-0)u}{2y}

dudx=(2y)u2y\frac{du}{dx}=\frac{(2y)u}{2y}

dudx=u\frac{du}{dx}=u

on solving we get integrated factor


u=exu=e^x

Multiply equation 1 by integrated factor

ex(x2+y2+2x)dx+2yexdy=0e^x(x^2+y^2+2x)dx+2ye^xdy=0

let M=(x2+y2+2x)ex(x^2+y^2+2x)e^x

N=2yex2ye^x

dMdy=2yex\frac{dM}{dy}=2ye^x

dNdx=2yex\frac{dN}{dx}=2ye^x


Hence dMdy=dNdx\frac{dM}{dy}=\frac{dN}{dx}


Solution of above equation is given by,


Let I(x,y) be a implict function

I(x,y)=M(x,y)dx\int M(x,y)dx

=ex(x2+y2+2x)dx\int e^x(x^2+y^2+2x)dx


=x2ex2xex+2ex+y2ex+2xex2exx^2e^x-2xe^x+2e^x+y^2e^x+2xe^x-2e^x

=x2ex+y2exx^2e^x+y^2e^x +f(y)


Let differentiate I(x,y) with respect to y

dIdy=\frac{dI}{dy}= N(x,y)

2yex+dfdy=2yex2ye^x+\frac{df}{dy}=2ye^x

dfdy=0=constant\frac{df}{dy}=0 =constant

I(x,y)=x2ex+y2ex+cx^2e^x+y^2e^x+c


at x=y=1x=y=1

0=e+e+cc+2e=0c=2e0=e+e+c\\ c+2e=0\\ c=-2e


hence this is the required solution is;

x2ex+y2ex2e=0x^2e^x+y^2e^x-2e=0

x2+y2=2e1xx^2+y^2=2e^{1-x} .



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