# Ordinary Differential Equations/Substitution 4

1)

${\displaystyle y'=csc(x+y)-1}$

${\displaystyle v=x+y}$

${\displaystyle v'=1+y'}$

${\displaystyle v'-1=csc(v)-1}$

${\displaystyle sin(v)dv=dx}$

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

${\displaystyle -cos(v)=x+C}$

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

${\displaystyle y=arccos(-x+C)-x}$

2)

${\displaystyle y'=csc({\frac {y}{x}})+{\frac {y}{x}}}$

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

${\displaystyle v'x+v=y'}$

${\displaystyle v'x+v=csc(v)+v}$

${\displaystyle v'x=csc(v)}$

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

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

${\displaystyle -cos(v)=ln(x)+C}$

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

${\displaystyle y=xarccos(-ln(x)+C)}$

3)

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

${\displaystyle v=sin(y^{2})}$

${\displaystyle v'=2yy'cos(y^{2})}$

${\displaystyle {\frac {v'}{2}}-v=0}$

${\displaystyle v'=2v}$

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

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

${\displaystyle ln(v)=2x+C}$

${\displaystyle v=Ce^{2x}}$

${\displaystyle sin(y^{2})=Ce^{2x}}$

${\displaystyle y^{2}=arcsin(Ce^{2x})}$

4)

${\displaystyle y'=yln(y)+y}$

${\displaystyle v=ln(y)}$

${\displaystyle v'={\frac {y'}{y}}}$

${\displaystyle v'y=yv+y}$

${\displaystyle v'=v+1}$

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

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

${\displaystyle ln(v+1)=x+C}$

${\displaystyle v+1=Ce^{x}}$

${\displaystyle v=Ce^{x}-1}$

${\displaystyle ln(y)=Ce^{x}-1}$

${\displaystyle y=e^{Ce^{x}-1}}$

5)

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

${\displaystyle v=x^{2}+y-1}$

${\displaystyle v'=2x+y'}$

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

${\displaystyle v'=v^{2}}$

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

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

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

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

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

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

6)

${\displaystyle y'={\frac {x^{2}}{y^{2}}}+{\frac {y}{x}}}$

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

${\displaystyle y'=v+xv'}$

${\displaystyle v+xv'={\frac {1}{v^{2}}}+v}$

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

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

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

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

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