Ordinary Differential Equations/Substitution 4

1)

$y^{\prime }=csc(x+y)-1$

$v=x+y$

$v^{\prime }=1+y^{\prime }$

$v^{\prime }-1=csc(v)-1$

$sin(v)dv=dx$

$\int sin(v)dv=\int dx$

$-cos(v)=x+C$

$-cos(x+y)=x+C$

$y=arccos(-x+C)-x$

2)

$y^{\prime }=csc(\frac{y}{x})+\frac{y}{x}$

$v=\frac{y}{x}$

$v^{\prime }x+v=y^{\prime }$

$v^{\prime }x+v=csc(v)+v$

$v^{\prime }x=csc(v)$

$sin(v)dv=\frac{dx}{x}$

$\int sin(v)dv=\int \frac{dx}{x}$

$-cos(v)=ln(x)+C$

$-cos(\frac{y}{x})=ln(x)+C$

$y=xarccos(-ln(x)+C)$

3)

$ycos(y^2)y^{\prime }-sin(y^2)=0$

$v=sin(y^2)$

$v^{\prime }=2y{y}^{\prime }cos(y^2)$

$\frac{v^{\prime }}{2}-v=0$

$v^{\prime }=2v$

$\frac{dv}{v}=2dx$

$\int \frac{dv}{v}=\int 2dx$

$ln(v)=2x+C$

$v=C{e}^{2x}$

$sin(y^2)=C{e}^{2x}$

$y^2=arcsin(C{e}^{2x})$

4)

$y^{\prime }=yln(y)+y$

$v=ln(y)$

$v^{\prime }=\frac{y^{\prime }}{y}$

$v^{\prime }y=yv+y$

$v^{\prime }=v+1$

$\frac{dv}{v+1}=dx$

$\int \frac{dv}{v+1}=\int dx$

$ln(v+1)=x+C$

$v+1=C{e}^x$

$v=C{e}^x-1$

$ln(y)=C{e}^x-1$

$y=e^{C{e}^x-1}$

5)

$y^{\prime }=(x^2+y-1{)}^2-2x$

$v=x^2+y-1$

$v^{\prime }=2x+y^{\prime }$

$2x+y^{\prime }=(x^2+y-1{)}^2$

$v^{\prime }=v^2$

$\frac{dv}{v^2}=dx$

$\int \frac{dv}{v^2}=\int dx$

$-\frac{1}{v}=x+C$

$v=\frac{-1}{x+C}$

$x^2+y-1=\frac{-1}{x+C}$

$y=\frac{-1}{x+C}-x^2+1$

6)

$y^{\prime }=\frac{x^2}{y^2}+\frac{y}{x}$

$v=\frac{y}{x}$

$y^{\prime }=v+x{v}^{\prime }$

$v+x{v}^{\prime }=\frac{1}{v^2}+v$

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

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

$\frac{1}{3}{v}^3=ln(x)+C$

$\frac{y^3}{3x^3}=ln(x)+C$

$y=(3x^3(ln(x)+C){)}^{\frac{1}{3}}$