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) $$y'=y^3sec^2(x)$$

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

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

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

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

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) $$y'=x^3/y^3$$

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

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

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

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

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

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

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

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

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

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

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

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