Answer to Question #212945 in Differential Equations for sasha

Question

1. solve each of the following

a) (6xy+2y²-5)dx +(3x²+4xy-6)dy=0 y(1)=2

b)x²d²y/dx²+xdy/dx+4y=2xlnx

Answer 1

a) $(6xy+2y^2-5)dx +(3x^2+4xy-6) dy=0$ , $y(1)=2$

$M(x,y)=6xy+2y^2-5 \implies {\partial M(x,y)\over \partial y}=M_y=6x+4y$

$N(x,y)=3x^2+4xy-6 \implies {\partial N(x,y)\over \partial x}=N_x=6x+4y$

Hence exact since $M_y=N_x$

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

${\partial \mu(x,y) \over \partial x}=M(x,y)=6xy+2y^2-5$

${\partial \mu(x,y) \over \partial y}=N(x,y)=3x^2+4xy-6$

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

$=\int^x(6xy+2y^2-5)dx$ $+g(y)$

$=3x^2y+2xy^2-5x$ $+g(y)$

$\implies {\partial \mu (x,y) \over \partial y}=3x^2+4xy+{ \partial g(y)\over \partial y}$

$\implies 3x^2+4xy+{ \partial g(y)\over \partial y}=3x^2+4xy-6$

$\implies { \partial g(y)\over \partial y}=-6$

$\implies \int dg(y)=-6 \int dy$

$\implies g(y)=-6y$

$\implies \mu(x,y)=3x^2y+2xy^2-5x-6y$

$\therefore$ The general solution is thus $3x^2y+2xy^2-5x-6y=c$

We can find the value of Cc using the initial condition $y(1)=2$

$\implies3(1)^2(2)+2(1)(2)^2-5(1)-6(2)=c \implies c=-3$

$\therefore$ the particular solution is

$3x^2y+2xy^2-5x-6y=-3$

Rearranging the equation gives

$3x^2y+2xy^2-5x+3=6y$

dividing through by 6 gives

$y={1\over 2}x^2y+{1\over 3}xy^2-{5\over 6}x+{1\over 2}$

b) $x^2{ d^2y\over dx^2}+x{dy\over dx}+4y=2xlnx$

This is a Cauchy's linear differential equation

let $x=e^z \implies z=lnx$

let $x{dy\over dx}=Dy$ , $D={d\over dz}$

let $x^2{ d^2y\over dx^2}=D(D-1)y$

therefore the given equation becomes

$=D(D-1)y+Dy+4y=2ze^z$

$=(D^2-D+D+4)y=2ze^z$

$=(D^2+4)y=2ze^z$

Next we need to find C.F as follows

$D^2+4=0\implies$ $D=2i$ and $D=-2i$ where $i$ is an imaginary number

$\implies C.F=C_1e^{-2iz}+C_2e^{2iz}=C_1sin2z+C_2cos2z$

but $z=log$ $x$

$\therefore$ $C.F=C_1sin 2logx+C_2cos2logx$

Next we need to find $P.I$ as follows

$P.I={1\over (D^2+4)}2ze^z=2e^z[{1\over (D+1)^2+4}\times z]=2e^z[{1\over D^2+2D+5}]\times z$

$={2e^z\over 5}[{1\over 1+{D^2+2D\over 5}}]z$

$={2e^z\over 5}[1+{D^2+2D\over 5}]^{-1}z$

$={2e^z\over 5}[1-{D^2+2D\over 5}+...]z$

$={2e^z\over 5}[z-{0^2+2\over 5}]$

$={2e^z\over 5}[z-{2\over 5}]$

$={2ze^z\over 5}-{4e^z\over 25}$

but $x=e^z$ and $z=lnx$

$\implies P.I={2xlogx\over 5}-{4logx\over 25}$

$\therefore$ The general solution is given as

$y=CF+PI$

$y=C_1sin 2logx+C_2cos2logx+{2xlogx\over 5}-{4logx\over 25}$