Answer to Question #108023 in Differential Equations for Peter Engalla Garcia

Question

Find a set of solutions for:
2(x-y)dx+(3x-y-1)dy=0

(2x+y-4)dx+(x-3y+12)dy=0

dy/dx=tanycotx - secycosx

Answer 1

1) Find a set of solutions for

$2(x-y)dx+(3x-y-1)dy=0\\$

$y'={2(y-x) \above{2pt} 3x-y-1}$

Solving the system

$y_ 0 ​ −x_ 0 ​ =0$

$3x_0-y_0-1=0$

$x_0={1 \above{2pt} 2}, y_0={1 \above{2pt} 2}$

We shall replace

$x=t+{1 \above{2pt} 2}$

$y=z+{1\above{2pt} 2}$

$y'=z'$

where t is a new independent variable,  z is a new function.Substitute into equation

$z'={2(z+ {1 \above{2pt} 2}-t- {1\above{2pt} 2}) \above{2pt} 3t+ {3 \above{2pt} 2} -z- {1 \above{2pt} 2}-1 }$

$z'={2(z-t) \above{2pt} 3t-z}$

we shall replace

${z \above{2pt} t}=u(t),z=t.u. z'=tu'+u$

then,

$tu'+u={2(tu-t) \above{2pt} 3t-tu}$

$tu'+u={2u-2\above{2pt} 3-u}$

$tu'={2u-2 \above{2pt} 3-u}-u$

$tu'={2u-2-3u+u^2 \above{2pt} 3-u}$

${3-u \above{2pt} (u+1)(u-2)} du={dt \above{2pt} t}$

$\int (-{4 \above{2pt} 3}{1 \above{2pt} u+1}+{1 \above{2pt} 3}{1 \above{2pt} u-2})du=\int {dt \above{2pt} t}$

$-{4 \above{2pt} 3}ln|u+1|+{1 \above{2pt} 3}ln|u-2|=ln|t|+{1 \above{2pt} 3} ln|c|$

${(u-2) \above{2pt} (u+1)^4}=ct^3 ,u+1=0$

where $u={y- {1\above{2pt} 2}\above{2pt} x-{1 \above{2pt} 2} }$

The solution of the equation are

${ {y- {1 \above{2pt} 2}\above{2pt} x- {1 \above{2pt} 2} } -2 \above{2pt}( {y- {1 \above{2pt} 2}\above{2pt} x- {1 \above{2pt} 2} } )^{4}+1 }$ $=c(x-{1 \above{2pt} 2})^{3}$

${y- {1\above{2pt} 2}\above{2pt} x-{1 \above{2pt} 2} }+1=0$

2) Find a set of solutions for

$(2x+y-4)dx+(x-3y+12)dy=0\\ y'=-\frac{2x+y-4}{x-3y+12}$

Solving the system

$2x_0+y_0-4=0$

$x _0-3y_0+12=0$

$x_0=0, y_0=4$

we shall replace

$x=t , y=z+4$

$y'=z'$

$z'=-{2t+z+4-4 \above{2pt} t-3z-12+12}$

$z'=-{2t +z\above{2pt} t-3z}$

we shall replace

${z \above{2pt} t}=u(t)$ ,$z=t.u, z'=tu'+u$

then

$tu'+u={2t+tu \above{2pt} 3tu-t}$

$tu'={2+u-3u^2+u \above{2pt} 3u-1}$

$-{3u-1 \above{2pt} 3u^2-2u-u} du={dt\above{2pt} t}$

$-{1 \above{2pt} 2}\int {d(3u^2-2u-2)\above{2pt} 3u^2-2u-2}du=\int{dt\above{2pt} t}$

$-{1\above{2pt} 2}ln|3u^2-2u-2|=ln|t|+{1 \above{2pt} 2}ln|c|$

${1 \above{2pt} 3u^2-2u-2}=ct^2$

$3u^2-2u-2=0$

where $u={y-4\above{2pt} x}$

Then solution of the above equation are

${1 \above{2pt} 3( {y-4 \above{2pt} x} )^{2}-2* {y-4\above{2pt} x}-2 }=cx^2$

$3({y-4 \above{2pt} x})^2-2*{y-4\above{2pt} x}-2=0$

3) Find a set of solutions for

$\frac{dy}{dx}=\tan y\cot x-\sec y\cos x\\ \frac{dy}{dx}=\frac{\sin y}{\cos y}\frac{\cos x}{\sin x}-\frac{1}{\cos y}\cos x\\ \frac{dy}{dx}=\frac{\sin y\cos x-\sin x \cos x}{\sin x\cos y}\\ (\sin x-\sin y)\cos x dx+\sin x \cos y dy=0$

$put$ $u=\sin y, du=\cos ydy$

Substituting, we get

$(sinx−u)cosxdx+sinxdu=0$

$(\sin x-u)\cos xdx+\sin x du=0\\ \frac{du}{dx}-u\cdot\frac{\cos x}{\sin x}=-\cos x$

The equation is first - order linear in u The integrating factor is

$I=e^{-\int\frac{\cos x}{\sin x}dx}=e^{-\ln\sin x}=\frac{1}{\sin x} ​$

Hence

$u\frac{1}{\sin x}=-\int\frac{\cos x}{\sin x}dx=-ln|\sin x|+c$

solve for u

$u=-\sin x\ln|\sin x|+c\sin x$

Put y back

$\sin y=-\sin x\ln|\sin x|+c\sin x$