Question

INTEGRATING FACTOR

Test for exactness and find the general solution:

1. y(x² + y)dx + x(x² − 2y)dy = 0

2. (2y⁴ + 1)dx + 4xy³dy = 0

3. y(x³y − 1)dx + x(xy⁴ − 1)dy = 0

4. y²dx + xydy = 0

5. y(y − 1)dx + xdy = 0

Answer 1

1. $y(x^2 + y)dx + x(x^2 − 2y)dy = 0$

$(x^2y + y^2)dx + (x^3 − 2xy)dy = 0$

$M(x,y)dx+N(x,y)dy=0$

$M(x,y)=x^2y+y^2\implies{\partial M(x,y)\over \partial y}=M_y =x^2 +2y$

$N(x,y)=x^3-2xy\implies{\partial N(x,y)\over \partial x}=N_y =3x^2 -2y$

hence not exact since $M_y \ne N_x$

Integrating Factor = $e^{ \int^x {M_y -N_x\over N}dx }=e^{ \int^x { x^2+2y-3x^2+2y\over x^3-2xy}dx }$

=$e^{ \int^x { -2(2y-x^2)\over x(2y-x^2)}dx}=e^{\int -{2\over x}dx}$

=$e^{-2\int {1\over x}dx}=e^{lnx^{-2}}={1\over x^2}$

multiplying the differential equation with the integrating factor ${1\over x^2}$ gives

${1\over x^2}[(x^2y + y^2)dx + (x^3 − 2xy)dy = 0]$

$[{x^2y+y^2\over x^2}]dx+[{x^3-2xy\over x^2}]dy=0$

$M(x,y)=y+{y^2\over x^2}\implies{\partial M(x,y)\over \partial y}=M_y =1+{2y\over x^2}$

$N(x,y)=x-{2y\over x}\implies{\partial N(x,y)\over \partial x}=N_y =1 +{2y\over x^2}$

hence exact since $M_y = N_x=1+{2y\over x^2}$

There exist a function $\mu(x,y)$ such that

${\partial \mu(x,y)\over \partial x}=M(x,y)=y+{y^2\over x^2}$ and

${\partial \mu(x,y)\over \partial y}=N(x,y)=x-{2y\over x}$

let $\mu(x,y)=\int^xM(x,y)dx+g(y)$

$=\int^x(y+{y^2\over x^2})dx+g(y)$

$\mu(x,y)=xy-{y^2\over x}+g(y)$

$\implies {\partial \mu(x,y)\over \partial y}=x-{2y\over x}+{\partial g(y)\over \partial y}$

$\implies x-{2y\over x}+{\partial g(y)\over \partial y}=x-{2y\over x}$

$\implies \int g (y)=\int 0dy \implies g(y)=0$

$\implies \mu(x,y)=xy-{y^2\over x}+0$

$\therefore$ The general solution is thus $xy-{y^2\over x}=c$

2. $(2y^4 + 1)dx + 4xy^3dy = 0$

$M(x,y)=2y^4+1\implies{\partial M(x,y)\over \partial y}=M_y =8y^3$

$N(x,y)=4xy^3\implies{\partial N(x,y)\over \partial x}=N_y =4y^3$

hence not exact since $M_y \ne N_x$

Integrating Factor = $e^{ \int^x {M_y -N_x\over N}dx }=e^{ \int^x { 8y^3-4y^3\over 4xy^3}dx }=e^{ \int^x { 1\over x}dx }=e^{ lnx}=x$

multiplying the differential equation with the integrating factor $x$ gives

$x[(2y^4 + 1)dx + 4xy^3dy = 0]$

$(2xy^4 + x)dx + 4x^2y^3dy = 0$

$M(x,y)=2xy^4+x\implies{\partial M(x,y)\over \partial y}=M_y =8xy^3$

$N(x,y)=4x^2y^3\implies{\partial N(x,y)\over \partial y}=N_y =8xy^3$

hence exact since $M_y = N_x=8xy^2$

There exist a function $\mu(x,y)$ such that

${\partial \mu(x,y)\over \partial x}=M(x,y)=2xy^4+x$ and

${\partial \mu(x,y)\over \partial y}=N(x,y)=4x^2y^3$

let $\mu(x,y)=\int^yN(x,y)dy+g(x)$

$=\int^y4x^2y^3dy+g(x)$

$\mu(x,y)=x^2y^4+g(x)$

$\implies {\partial \mu(x,y)\over \partial x}=2xy^4+{\partial g(x)\over \partial x}$

$\implies 2xy^4+{\partial g(x)\over \partial x}=2xy^4+x$

$\implies \int d g (x)=\int xdx \implies g(x)={x^2\over 2}$

$\implies \mu(x,y)=x^2y^4+{x^2\over 2}$

$\therefore$ The general solution is thus $x^2y^4+{x^2\over 2}=c$

3.$y(x^3y − 1)dx + x(xy^4 − 1)dy = 0$

$(x^3y ^2− y)dx + (x^2y^4 − x)dy = 0$

$M(x,y)=x^3y ^2− y\implies{\partial M(x,y)\over \partial y}=M_y =2x^3y-1$

$N(x,y)=x^2y^4 − x\implies{\partial N(x,y)\over \partial x}=N_x =2xy^4-1$

hence not exact since $M_y \ne N_x$

The integrating factor of the form p(x) or p(y) do not exist for the differential equation hence has no solution

4.$y^2dx + xydy = 0$

$M(x,y)=y^2\implies{\partial M(x,y)\over \partial y}=M_y =2y$

$N(x,y)=xy\implies{\partial N(x,y)\over \partial x}=N_x =y$

hence not exact since $M_y \ne N_x$

Integrating Factor = $e^{ \int^x {M_y -N_x\over N}dx }=e^{ \int^x { 2y-y\over xy}dx }=e^{ \int^x { 1\over x}dx }=e^{ lnx}=x$

multiplying the differential equation with the integrating factor $x$ gives

$x[ y^2dx + xydy = 0]$

$xy^2dx + x^2ydy = 0$

$M(x,y)=xy^2\implies{\partial M(x,y)\over \partial y}=M_y =2xy$

$N(x,y)=x^2y\implies{\partial N(x,y)\over \partial x}=N_x =2xy$

hence exact since $M_y = N_x=2xy$

There exist a function $\mu(x,y)$ such that

${\partial \mu(x,y)\over \partial x}=M(x,y)=xy^2$ and

${\partial \mu(x,y)\over \partial y}=N(x,y)=x^2y$

let $\mu(x,y)=\int^xM(x,y)dx+g(y)$

$=\int^xxy^2dy+g(x)$

$\mu(x,y)={x^2\over 2}y^2+g(x)$

$\implies {\partial \mu(x,y)\over \partial y}=x^2y+{\partial g(y)\over \partial y}$

$\implies x^2y+{\partial g(y)\over \partial y}=x^2y$

$\implies \int d g (y)=\int 0dx \implies g(x)=0$

$\implies \mu(x,y)={1\over 2}x^2y^2+0$

$\therefore$ The general solution is thus ${1\over 2}x^2y^2=c$

5. $y(y − 1)dx + xdy = 0$

$(y^2 − y)dx + xdy = 0$

$M(x,y)=y^2-y\implies{\partial M(x,y)\over \partial y}=M_y =2y-1$

$N(x,y)=x\implies{\partial N(x,y)\over \partial x}=N_x =1$

hence not exact since $M_y \ne N_x$

Integrating Factor = $e^{ \int^y {M_y -N_x\over N}dy }=e^{ \int^y { 1-2y+1\over y^2-y}dy }=e^{ \int^y{ 2-2y\over y^2-y}dy }$

evaluate ${ \int^y{ 2-2y\over y^2-y}dy }$ by partial fraction

${ 2-2y\over y(y-1)}={A\over y}+{B\over y-1}$

${ 2-2y\over y(y-1)}={A(y-1)+By\over y(y-1)}$

$\implies 2-2y=Ay-A+By=(A+B)y-A$

equating coefficients of y terms together and constants together we have

$A+B=-2 and A=-2\implies B=0$

$\therefore{ 2-2y\over y(y-1)}={-2\over y}+{0\over y-1}={-2\over y}$

$\therefore { \int^y{ 2-2y\over y^2-y}dy }=-2\int{1\over y}dy=-2lny=ln y^{-2}$

$\therefore$ Integrating Factor $=e^{lny^{-2}}={1\over y^2}$

multiplying the differential equation with the integrating factor ${1\over y^2}$ gives

${1\over y^2}[ (y^2 − y)dx + xdy = 0]$

${y-1\over y}dx + {x\over y^2}dy = 0$

$M(x,y)={y-1\over y}\implies{\partial M(x,y)\over \partial y}=M_y ={1\over y^2}$

$N(x,y)={x\over y^2}\implies{\partial N(x,y)\over \partial x}=N_x ={1\over y^2}$

hence exact since $M_y = N_x={1\over y^2}$

There exist a function $\mu(x,y)$ such that

${\partial \mu(x,y)\over \partial x}=M(x,y)={y-1\over y}$ and

${\partial \mu(x,y)\over \partial y}=N(x,y)={x\over y^2}$

let $\mu(x,y)=\int^xM(x,y)dx+g(y)$

$=\int^x{y-1\over y}dx+g(y)$

$\mu(x,y)=x{y-1\over y}+g(y)$

$\implies {\partial \mu(x,y)\over \partial y}={x\over y^2}+{\partial g(y)\over \partial y}$

$\implies {x\over y^2}+{\partial g(y)\over \partial y}={x\over y^2}$

$\implies \int d g (y)=\int 0dy \implies g(y)=0$

$\implies \mu(x,y)=x{y-1\over y}+0$

$\therefore$ The general solution is thus $x{y-1\over y}=c$