# Question of Differential Equations

• Aug 31st 2010, 06:15 AM
Paymemoney
Question of Differential Equations
1)Solve $y' = \sqrt{1-y^2}$
$y=sin(u)$ $dy=cos(u) dy$
$y=\int \sqrt{1-y^2}$
$y=cos(u)$

what should i do next if this is correct?

2)Find the orthogonal trajectories for $y=Ce^{x}$
$\frac{dy}{dx}=e^x$
$m=e^x$
O.T $= \int \frac{-1}{e^x}$
O.T $= -e^{-x}+C$

book's answer is $y^2=k-2x$

P.S
• Aug 31st 2010, 06:44 AM
Prove It
$\frac{dy}{dx} = \sqrt{1 - y^2}$

$\frac{1}{\sqrt{1 - y^2}}\,\frac{dy}{dx} = 1$

$\int{\frac{1}{\sqrt{1 - y^2}}\,\frac{dy}{dx}\,dx} = \int{1\,dx}$

$\int{\frac{1}{\sqrt{1 - y^2}}\,dy} = x + C_1$

Now make the substitution $y = \sin{\theta}$ so that $\theta = \arcsin{y}$ and $dy = \cos{\theta}\,d\theta$, then the left hand integral becomes

$\int{\frac{1}{\sqrt{1 - \sin^2{\theta}}}\,\cos{\theta}\,d\theta} = x + C_1$

$\int{\frac{\cos{\theta}}{\cos{\theta}}\,d\theta} = x + C_1$

$\int{1\,d\theta} = x + C_1$

$\theta + C_2 = x + C_1$

$\arcsin{y} = x + C$ where $C = C_1 - C_2$

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

$y = \sin{x}\cos{C} + \cos{x}\sin{C}$

$y = A\sin{x} + B\cos{x}$, where $A = \cos{C}$ and $B = \sin{C}$.
• Aug 31st 2010, 04:01 PM
Paymemoney
any ideas how the second question should be done?
• Aug 31st 2010, 04:57 PM
Jester
Quote:

Originally Posted by Paymemoney

2)Find the orthogonal trajectories for $y=Ce^{x}$
$\frac{dy}{dx}=e^x$
$m=e^x$
O.T $= \int \frac{-1}{e^x}$
O.T $= -e^{-x}+C$

book's answer is $y^2=k-2x$

P.S

If $y = C e^x$ then $y e^{-x} = C$ so $y'e^{-x} - y e^{-x} = 0 \;\; \text{or}\;\; y' = y$. Your orthogonal trajectories satisfies $y' \cdot y'_{\perp} = -1$

then

$y'_{\perp} = - \dfrac{1}{y}$ (you can integrate this giving your result).
• Aug 31st 2010, 06:09 PM
Paymemoney
thanks