Variable Separable & Homogeneous Differential Equations

Explanation:

This is a homogeneous differential equation of degree 0. Substitute $y=vx⟹{\displaystyle \frac{dy}{dx}}=v+x{\displaystyle \frac{dv}{dx}}$.

Substitute this into the original equation:
$v+x{\displaystyle \frac{dv}{dx}}={\displaystyle \frac{vx-x}{vx+x}}={\displaystyle \frac{v-1}{v+1}}$

$x{\displaystyle \frac{dv}{dx}}={\displaystyle \frac{v-1}{v+1}}-v$
$x{\displaystyle \frac{dv}{dx}}={\displaystyle \frac{v-1-v^2-v}{v+1}}=-{\displaystyle \frac{1+v^2}{v+1}}$

Separate the variables:
${\displaystyle \frac{v+1}{1+v^2}}dv=-{\displaystyle \frac{dx}{x}}$

Integrate both sides:
$\int [{\displaystyle \frac{v}{1+v^2}}+{\displaystyle \frac{1}{1+v^2}}]dv=-\int {\displaystyle \frac{dx}{x}}$
${\displaystyle \frac{1}{2}}\ln (1+v^2)+\arctan (v)=-\ln |x|+C$

Apply the initial condition $y(1)=1⟹$ at $x=1,v=1$:
${\displaystyle \frac{1}{2}}\ln (1+1)+\arctan (1)=-\ln (1)+C$
$C={\displaystyle \frac{1}{2}}\ln 2+{\displaystyle \frac{\pi }{4}}$

Substitute $v={\displaystyle \frac{y}{x}}$ back into the general solution:
${\displaystyle \frac{1}{2}}\ln (1+{\displaystyle \frac{y^2}{x^2}})+\arctan \left({\displaystyle \frac{y}{x}}\right)+\ln |x|={\displaystyle \frac{1}{2}}\ln 2+{\displaystyle \frac{\pi }{4}}$
${\displaystyle \frac{1}{2}}\ln \left({\displaystyle \frac{x^2+y^2}{x^2}}\right)+\ln |x|+\arctan \left({\displaystyle \frac{y}{x}}\right)={\displaystyle \frac{1}{2}}\ln 2+{\displaystyle \frac{\pi }{4}}$
${\displaystyle \frac{1}{2}}\ln (x^2+y^2)-{\displaystyle \frac{1}{2}}\ln (x^2)+\ln |x|+\arctan \left({\displaystyle \frac{y}{x}}\right)={\displaystyle \frac{1}{2}}\ln 2+{\displaystyle \frac{\pi }{4}}$
Since ${\displaystyle \frac{1}{2}}\ln (x^2)=\ln |x|$, the terms cancel out.

Final Solution: ${\displaystyle \frac{1}{2}}\ln (x^2+y^2)+\arctan \left({\displaystyle \frac{y}{x}}\right)={\displaystyle \frac{1}{2}}\ln 2+{\displaystyle \frac{\pi }{4}}$

Explanation:

This equation can be simplified using substitution. Let $v=x+y$.
Differentiating with respect to $x$: ${\displaystyle \frac{dv}{dx}}=1+{\displaystyle \frac{dy}{dx}}⟹{\displaystyle \frac{dy}{dx}}={\displaystyle \frac{dv}{dx}}-1$.

Substitute back into the equation:
${\displaystyle \frac{dv}{dx}}-1=\sin (v)+\cos (v)$
${\displaystyle \frac{dv}{dx}}=1+\sin (v)+\cos (v)$

Separate the variables:
${\displaystyle \frac{dv}{1+\sin (v)+\cos (v)}}=dx$

Use the half-angle formulas: $1+\cos (v)=2{\cos }^2\left({\displaystyle \frac{v}{2}}\right)$ and $\sin (v)=2\sin \left({\displaystyle \frac{v}{2}}\right)\cos \left({\displaystyle \frac{v}{2}}\right)$.
${\displaystyle \frac{dv}{2{\cos }^2(v/2)+2\sin (v/2)\cos (v/2)}}=dx$
${\displaystyle \frac{dv}{2{\cos }^2(v/2)[1+\tan (v/2)]}}=dx$
${\displaystyle \frac{\frac{1}{2}{\sec }^2(v/2)dv}{1+\tan (v/2)}}=dx$

Let $t=\tan (v/2)$, then $dt={\displaystyle \frac{1}{2}}{\sec }^2(v/2)dv$.
$\int {\displaystyle \frac{dt}{1+t}}=\int dx$
$\ln |1+t|=x+C$

Substitute back $t$ and then $v$:
Final Solution: $\ln |1+\tan \left({\displaystyle \frac{x+y}{2}}\right)|=x+C$

Explanation:

Rewrite the equation to isolate ${\displaystyle \frac{dy}{dx}}$:
$x^2{\displaystyle \frac{dy}{dx}}=-y(x+y)$
${\displaystyle \frac{dy}{dx}}=-{\displaystyle \frac{y(x+y)}{x^2}}=-({\displaystyle \frac{y}{x}}+{\displaystyle \frac{y^2}{x^2}})$

This is a homogeneous equation. Substitute $y=vx⟹{\displaystyle \frac{dy}{dx}}=v+x{\displaystyle \frac{dv}{dx}}$.
$v+x{\displaystyle \frac{dv}{dx}}=-v-v^2$
$x{\displaystyle \frac{dv}{dx}}=-2v-v^2=-v(v+2)$

Separate the variables:
${\displaystyle \frac{dv}{v(v+2)}}=-{\displaystyle \frac{dx}{x}}$

Use partial fractions for the left side:
${\displaystyle \frac{1}{v(v+2)}}={\displaystyle \frac{1}{2}}[{\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{v+2}}]$

Integrate both sides:
${\displaystyle \frac{1}{2}}\int ({\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{v+2}})dv=-\int {\displaystyle \frac{dx}{x}}$
${\displaystyle \frac{1}{2}}\ln |v|-{\displaystyle \frac{1}{2}}\ln |v+2|=-\ln |x|+C_1$
${\displaystyle \frac{1}{2}}\ln \left|{\displaystyle \frac{v}{v+2}}\right|=-\ln |x|+C_1$
$\ln \left|{\displaystyle \frac{v}{v+2}}\right|=-2\ln |x|+2C_1=\ln (x^{-2})+C_2$
${\displaystyle \frac{v}{v+2}}={\displaystyle \frac{K}{x^2}}$ (where $K=e^{C_2}$)

Substitute back $v={\displaystyle \frac{y}{x}}$:
${\displaystyle \frac{y/x}{y/x+2}}={\displaystyle \frac{K}{x^2}}$
${\displaystyle \frac{y}{y+2x}}={\displaystyle \frac{K}{x^2}}$

Final Solution: $x^{2}y=K(2x+y)$

Explanation:

This is a differential equation reducible to a homogeneous form. First, find the intersection of the two lines $x+2y-3=0$ and $2x+y-3=0$. Solving these simultaneously yields the point $(1,1)$.

Perform a coordinate shift: let $X=x-1$ and $Y=y-1$. Then $dX=dx$ and $dY=dy$.
The equation becomes homogeneous:
${\displaystyle \frac{dY}{dX}}={\displaystyle \frac{X+2Y}{2X+Y}}$

Substitute $Y=vX⟹{\displaystyle \frac{dY}{dX}}=v+X{\displaystyle \frac{dv}{dX}}$.
$v+X{\displaystyle \frac{dv}{dX}}={\displaystyle \frac{X+2vX}{2X+vX}}={\displaystyle \frac{1+2v}{2+v}}$
$X{\displaystyle \frac{dv}{dX}}={\displaystyle \frac{1+2v}{2+v}}-v={\displaystyle \frac{1+2v-2v-v^2}{2+v}}={\displaystyle \frac{1-v^2}{2+v}}$

Separate the variables:
${\displaystyle \frac{2+v}{1-v^2}}dv={\displaystyle \frac{dX}{X}}$

Split the integral on the left:
$\int {\displaystyle \frac{2}{1-v^2}}dv+\int {\displaystyle \frac{v}{1-v^2}}dv=\int {\displaystyle \frac{dX}{X}}$
$2[{\displaystyle \frac{1}{2}}\ln \left|{\displaystyle \frac{1+v}{1-v}}\right|]-{\displaystyle \frac{1}{2}}\ln |1-v^2|=\ln |X|+C$
$\ln \left|{\displaystyle \frac{1+v}{1-v}}\right|-\ln \sqrt{1-v^2}=\ln |X|+C$

Using logarithm properties:
$\ln \left|{\displaystyle \frac{1+v}{(1-v)\sqrt{(1-v)(1+v)}}}\right|=\ln |X|+C$
$\ln \left|{\displaystyle \frac{\sqrt{1+v}}{(1-v{)}^{3/2}}}\right|=\ln |X|+C$
${\displaystyle \frac{1+v}{(1-v{)}^3}}=K{X}^2$

Substitute back $v={\displaystyle \frac{Y}{X}}$:
${\displaystyle \frac{1+Y/X}{(1-Y/X{)}^3}}=K{X}^2$
${\displaystyle \frac{X^2(X+Y)}{(X-Y{)}^3}}=K{X}^2⟹(X+Y)=K(X-Y{)}^3$
Substitute back $X=x-1$ and $Y=y-1$:
$(x-1+y-1)=K(x-1-y+1{)}^3$
$(x+y-2)=K(x-y{)}^3$

Final Solution: $(x-y{)}^3=C(x+y-2)$
(Note: C is the reciprocal of K from the derivation).