Differential Equations/Separable 4

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[edit] Existance 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.

[edit] Separable equations

7) $y' = y 3 s e c 2 (x)$

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

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

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

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

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) $y' = x 3 / y 3$

$y 3 d y = x 3 d x$

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

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

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

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

$d y = (x 2 + 3 x - 9) d x$

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

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

11) $y' = c o s (y) / s i n (y)$

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

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

$- l n (c o s (y)) = x + C$

$y = a r c c o s (C e x)$

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

$s i n (y) d y = c o s (x) d x$

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

$- c o s (y) = s i n (x) + C$

$y = a r c c o s ( - s i n (x) + C)$

[edit] Initial value problems

13) $y' = c o s (x) + s i n (x), y (0) = 1$

$d y = (c o s (x) + s i n (x)) d x$

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

$y = s i n (x) - c o s (x) + C$

$1 = s i n (0) - c o s (0) + C = 0 - 1 + C = C - 1$

$C = 2$

$y = s i n (x) - c o s (x) + 2$

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

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

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

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

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

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

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

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

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Category: Differential Equations

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