Answer to Question #212951 in Differential Equations for satek

Question

find the general solutions for each of the following

1. (3x²+9xy+5y²)dx-(6x²+4xy)dy =0 y(2)=-6
2. x²dy/dx +xy=y³/x

Answer 1

(1) Given equation is

$(3x^2+9xy+5y^2)dx-(6x^2+4xy)dy =0$

$(3x^2+9xy+5y^2)dx=(6x^2+4xy)dy$

$\frac{dy}{dx}=\frac{(3x^2+9xy+5y^2)}{(6x^2+4xy)}$

Putting y=vx, then $\frac{dy}{dx}=v+x\frac{dv}{dx}$

Then equation will be,

$v+x\frac{dv}{dx}=\frac{(3x^2+9vx^2+5v^2x^2)}{(6x^2+4vx^2)}$

$v+x\frac{dv}{dx}=\frac{(3+9v+5v^2)}{(6+4v)}$

$x\frac{dv}{dx}=\frac{(3+9v+5v^2)}{(6+4v)}-v$

$x\frac{dv}{dx}=\frac{(3+3v+v^2)}{(6+4v)}$

$\frac{(6+4v)dv}{(3+3v+v^2)}=\frac{dx}{x}$

Integrating both sides,

Integrating left hand side first,

$\int \frac{(6+4v)dv}{(3+3v+v^2)}$

Let $v^2+3v+3 = t$

$(2v+3)dv = dt$

Then integral will be,

$\int \frac{(6+4v)dv}{(3+3v+v^2)} = \int \frac{2 dt}{t} = 2ln|t| = 2ln|3+3v+v^2|$

Then, $2ln|3+3v+v^2| = ln|x| + ln|C|$

$(3+3v+v^2)^2 = Cx$

Putting v=y/x,

$(3+3(\frac{y}{x})+(\frac{y}{x})^2)^2 = Cx$

$(3x^2+3xy+y^2)^2 = Cx^5$ is the solution of the equation.

(2) Given equation is, $x^2\frac{dy}{dx} +xy=\frac{y^3}{x}$

Then, $\frac{dy}{dx} +\frac{y}{x}=\frac{y^3}{x^3}$

Putting y=vx, we will get,

$v+x\frac{dy}{dx}+v = v^3$

$x\frac{dv}{dx} = v^3-2v$

$\frac{dv}{ v^3-2v} =\frac{dx}{x}$

Integrating both sides,

Integrating left-hand side first, $\int\frac{dv}{ v^3-2v}$

$\int\frac{dv}{ v^3-2v} = \int\frac{dv}{ v(v^2-2)}$

Doing partial fraction,

$\frac{1}{ v(v^2-2)} = \frac{A}{v}+\frac{Bv+C}{v^2-2}$

Solving For A, B and C, we get

$\frac{1}{ v(v^2-2)} = \frac{A(v^2-2)+Bv^2+Cv}{v(v^2-2)} = \frac{(A+B)v^2+Cv-2A}{v(v^2-2)}$

Comparing powers on both sides, we get

A+B=0, C=0 and -2A=1

$\implies A = -\frac{1}{2} , B = \frac{1}{2} , C=0$

Then,

$\frac{1}{ v(v^2-2)} = \frac{-1}{2v}+\frac{v}{2(v^2-2)}$

Now integrating it,

$\int\frac{1}{ v(v^2-2)} dv=\int \frac{-1}{2v}+\frac{v}{2(v^2-2)} dv$

$\int \frac{v}{2(v^2-2)} dv =\int \frac{2v}{4(v^2-2)} dv$

Taking $v^2-2 =t \implies 2vdv = dt$

$\int \frac{2v}{4(v^2-2)} dv = \int \frac{dt}{4t} = \frac{1}{4}ln|t|= \frac{1}{4}ln|v^2-2|$

So, integral of LHS will be,

$\int\frac{1}{ v(v^2-2)} dv=\int \frac{-1}{2v}+\frac{v}{2(v^2-2)} dv = -\frac{1}{2}ln|v|+\frac{1}{4}ln|v^2-2|$

So, we get,

$-\frac{1}{2}ln|v|+\frac{1}{4}ln|v^2-2| = ln|x|+ln|C|$

$-\frac{2}{4}ln|v|+\frac{1}{4}ln|v^2-2| = ln|x|+ln|C|$

$\frac{1}{4}ln|\frac{v^2-2}{v^2}| = ln|x| + ln|C|$

$\frac{v^2-2}{v^2} = Kx^4$ where K is also some constant.

putting $v = \frac{y}{x}$

$\frac{y^2-2x^2}{y^2} = Kx^4$ which is the required solution.