Ordinary Differential Equations/Substitution 4

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1)

y'=csc(x+y)-1

v=x+y

v'=1+y'

v'-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'=csc(\frac{y}{x})+\frac{y}{x}

v=\frac{y}{x}

v'x+v=y'

v'x+v=csc(v)+v

v'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=x arccos(-ln(x)+C)



3)

ycos(y^2)y'-sin(y^2)=0

v=sin(y^2)

v'=2yy'cos(y^2)

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

v'=2v

\frac{dv}{v}=2dx

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

ln(v)=2x+C

v=Ce^{2x}

sin(y^2)=Ce^{2x}

y^2=arcsin(Ce^{2x})



4)

y'=yln(y)+y

v=ln(y)

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

v'y=yv+y

v'=v+1

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

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

ln(v+1)=x+C

v+1=Ce^x

v=Ce^x-1

ln(y)=Ce^x-1

y=e^{Ce^x-1}



5)

y'=(x^2+y-1)^2-2x

v=x^2+y-1

v'=2x+y'

2x+y'=(x^2+y-1)^2

v'=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'=\frac{x^2}{y^2}+\frac{y}{x}

v=\frac{y}{x}

y'=v+xv'

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

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

\int v^2dv=\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}}