# Ordinary Differential Equations/Separable 4

## Existence problems

1) f(x,y) has no discontinuities, so a solution exists. ${\displaystyle {\frac {\partial {f}}{\partial {y}}}}$ has no discontinuities, so the solution is unique.

2) f(x,y) is not defined for the point (-1,10) because ln(x) is not defined. So no solution exists.

3) f(x,y) has discontinuities at y=1 and -1, but not at 0 so a solution exists. ${\displaystyle {\frac {\partial {f}}{\partial {y}}}}$ has no discontinuities at (0,16) so the solution is unique.

4) f(x,y) has discontinuities at y<0, but not at 1 so a solution exists. ${\displaystyle {\frac {\partial {f}}{\partial {y}}}}$ is discontinuous at 1, so the solution is not unique

5) f(x,y) has discontinuities at -3 and -4, but not at 0 so a solution exists. ${\displaystyle {\frac {\partial {f}}{\partial {y}}}}$ has no discontinuities at (5,9) so the solution is unique.

6) f(x,y) has a discontinuity at x=5, so no solution exists.

## Separable equations

7) ${\displaystyle y'=y^{3}sec^{2}(x)}$

${\displaystyle {\frac {dy}{y^{3}}}=sec^{2}(x)dx}$

${\displaystyle \int {\frac {dy}{y^{3}}}=\int sec^{2}(x)dx}$

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

${\displaystyle y=-{\frac {1}{\sqrt {(2tan(x)+C)}}}}$

8) ${\displaystyle y'={\frac {5y^{2}+6}{y}}}$

${\displaystyle {\frac {ydy}{5y^{2}+6}}=dx}$

${\displaystyle \int {\frac {ydy}{5y^{2}+6}}=\int dx}$

${\displaystyle {\frac {1}{10}}ln(5y^{2}+6)=x+C}$

${\displaystyle y=\pm {\sqrt {Ce^{10x}-{\frac {6}{5}}}}}$

9) ${\displaystyle y'=x^{3}/y^{3}}$

${\displaystyle y^{3}dy=x^{3}dx}$

${\displaystyle \int y^{3}dy=\int x^{3}dx}$

${\displaystyle {\frac {1}{4}}y^{4}={\frac {1}{4}}x^{4}+C}$

${\displaystyle y=(x^{4}+C)^{\frac {1}{4}}}$

10) ${\displaystyle y'=x^{2}+3x-9}$

${\displaystyle dy=(x^{2}+3x-9)dx}$

${\displaystyle \int dy=\int (x^{2}+3x-9)dx}$

${\displaystyle y={\frac {1}{3}}x^{3}+{\frac {3}{2}}x^{2}-9x+C}$

11) ${\displaystyle y'=cos(y)/sin(y)}$

${\displaystyle {\frac {sin(y)dy}{cos(y)}}=dx}$

${\displaystyle \int {\frac {sin(y)dy}{cos(y)}}=\int dx}$

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

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

12) ${\displaystyle y'={\frac {cos(x)}{sin(y)}}}$

${\displaystyle sin(y)dy=cos(x)dx}$

${\displaystyle \int sin(y)dy=\int cos(x)dx}$

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

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

## Initial value problems

13) ${\displaystyle y'=cos(x)+sin(x),y(0)=1}$

${\displaystyle dy=(cos(x)+sin(x))dx}$

${\displaystyle \int dy=\int (cos(x)+sin(x))dx}$

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

${\displaystyle 1=sin(0)-cos(0)+C=0-1+C=C-1}$

${\displaystyle C=2}$

${\displaystyle y=sin(x)-cos(x)+2}$

14) ${\displaystyle y'=7y^{2},y(5)=9}$

${\displaystyle {\frac {dy}{y^{2}}}=7dx}$

${\displaystyle \int {\frac {dy}{y^{2}}}=\int 7dx}$

${\displaystyle -{\frac {1}{y}}=7x+C}$

${\displaystyle y={\frac {1}{-7x+C}}}$

${\displaystyle 9={\frac {1}{-7*5+C}}}$

${\displaystyle C={\frac {316}{9}}}$

${\displaystyle y={\frac {1}{-7x+{\frac {316}{9}}}}}$