Answer to Question #132210 in Differential Equations for Rouhish Ray

Question

Show that the Integrating factor of the equation (x^2+y^2+2x)dx + 2ydy=0 is e^x and its particular solution is x^2 + y^2=2e^1-x when x=y=1

Answer 1

We have the differential equation:

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

let f(x,y)=$x^2+y^2+2x$

g(y)=2y

so

f'$_y(x,y)=2y$

g'(x)=0

using Integration factor method,

$\frac{du}{dx}=\frac{(f_y-g_x)u}{g}$

$\frac{du}{dx}=\frac{(2y-0)u}{2y}$

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

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

on solving we get integrated factor

u=e$^x$

Multiply equation 1 by integrated factor

$e^x(x^2+y^2+2x)dx+2ye^xdy=0$

let M=$(x^2+y^2+2x)e^x$

N=$2ye^x$

$\frac{dM}{dy}=2ye^x$

$\frac{dN}{dx}=2ye^x$

Hence $\frac{dM}{dy}=\frac{dN}{dx}$

Solution of above equation is given by,

Let I(x,y) be a implict function

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

=$\int e^x(x^2+y^2+2x)dx$

=$x^2e^x-2xe^x+2e^x+y^2e^x+2xe^x-2e^x$

=$x^2e^x+y^2e^x$ +f(y)

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

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

$2ye^x+\frac{df}{dy}=2ye^x$

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

I(x,y)=$x^2e^x+y^2e^x+c$

at x=y=1

0=e+e+c

c+2e=0

c=-2e

so Particular solution becomes

$x^2e^x+y^2e^x-2e=0$

$x^2+y^2=2e^{1-x}$ .

This is the required answer.