Answer to Question #86910 in Differential Equations for Swati malik

Question

Solve the differential equation.
1. dy/dx=cot(y+x)-1
2. (x^2+y^2)dy/dx=xy.

Answer 1

Solution:

1.If

$\frac{dy}{dx}=cot(x+y)-1$

then we make a replacement:

$x+y=z,y=z-x,\frac{dy}{dx}=\frac{dz}{dx}-1$

and have

$\frac{dz}{dx}-1=cot(z)-1,\int{\frac{dz}{cot(z)}}=\int{dx},$

$\int{\frac{sin(z)dz}{cos(z)}}=\int{dx},-ln|cos(z)|=x+c,|cos(z)|=e^{-x-c},$

$|cos(x+y)|=e^{-x-c},$ $y=arccos(e^{-x-c}) -x,-\frac{\pi}{2}+2\pi n\le{x}\le{\frac{\pi}{2}}+2\pi n,n\isin{Z}$

and

$y=arccos(-e^{-x-c}) -x,(\frac{\pi}{2}+2\pi n\le{x}\le{\frac{3\pi+2\pi n}{2}})n\isin{Z}.$

2.If

$(x^2+y^2)\frac{dy}{dx}=xy$

then we make a replacement:

$\frac{y}{x}=z,y=zx,\frac{dx}{dy}=\frac{dz}{dx}x+z$

and have

$(x^2+x^2z^2)(\frac{dz}{dx}x+z)=x^2z$ $\frac{1+z^2}{-z^3}\frac{dz}{dx}x=1,\int{(-\frac{1}{z^3}-\frac{1}{z})dz}=\int{\frac{dx}{x}},$

$\frac{1}{2z^2}-ln|z|=ln|x|+ln|c|,\frac{x^2}{2y^2}-ln|y|=ln|c|,$

$x^2=y^2ln(cy)^2.$

Answer:1.

$y=arccos(e^{-x-c}) -x,-\frac{\pi}{2}+2\pi n\le{x}\le{\frac{\pi}{2}}+2\pi n,n\isin{Z}$

and

$y=arccos(-e^{-x-c}) -x,(\frac{\pi}{2}+2\pi n\le{x}\le{\frac{3\pi+2\pi n}{2}})n\isin{Z}.$

2.

$x^2=y^2ln(cy)^2.$