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) $y^{\prime }=y^{3}se{c}^2(x)$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

11) $y^{\prime }=cos(y)/sin(y)$

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

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

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

$y=arccos(C{e}^x)$

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

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

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

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

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

Initial value problems

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

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

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

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

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

$C=2$

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

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

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

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

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

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

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

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

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