Ordinary Differential Equations/Separable 4

Existence problems

1) f(x,y) has no discontinuities, so a solution exists. ${\frac {\partial {f}}{\partial {y}}}$ 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. ${\frac {\partial {f}}{\partial {y}}}$ 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. ${\frac {\partial {f}}{\partial {y}}}$ 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. ${\frac {\partial {f}}{\partial {y}}}$ 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) $y'=y^{3}sec^{2}(x)$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

11) $y'=cos(y)/sin(y)$

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

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

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

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

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

$sin(y)dy=cos(x)dx$

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

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

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

Initial value problems

13) $y'=cos(x)+sin(x),y(0)=1$

$dy=(cos(x)+sin(x))dx$

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

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

$1=sin(0)-cos(0)+C=0-1+C=C-1$

$C=2$

$y=sin(x)-cos(x)+2$

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

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

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

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

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

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

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

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