# Question of Differential Equations

• Aug 31st 2010, 05:15 AM
Paymemoney
Question of Differential Equations
1)Solve $\displaystyle y' = \sqrt{1-y^2}$
$\displaystyle y=sin(u)$$\displaystyle dy=cos(u) dy$
$\displaystyle y=\int \sqrt{1-y^2}$
$\displaystyle y=cos(u)$

what should i do next if this is correct?

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

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

P.S
• Aug 31st 2010, 05:44 AM
Prove It
$\displaystyle \frac{dy}{dx} = \sqrt{1 - y^2}$

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

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

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

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

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

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

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

$\displaystyle \theta + C_2 = x + C_1$

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

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

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

$\displaystyle y = A\sin{x} + B\cos{x}$, where $\displaystyle A = \cos{C}$ and $\displaystyle B = \sin{C}$.
• Aug 31st 2010, 03:01 PM
Paymemoney
any ideas how the second question should be done?
• Aug 31st 2010, 03:57 PM
Jester
Quote:

Originally Posted by Paymemoney

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

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

P.S

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

then

$\displaystyle y'_{\perp} = - \dfrac{1}{y}$ (you can integrate this giving your result).
• Aug 31st 2010, 05:09 PM
Paymemoney
thanks