Am i doing this right

1. Feb 8, 2006

jamesbob

Hey, just wanna check iv done these questions right so far...

$$\mbox{Find the general solution to} \frac{dy}{dx} = y^2xcos(2x) \mbox{giving explicitly in terms of x.}$$
$$\mbox{Find the particular solution satisfying y(0) = -1}$$

$$\frac{dy}{y^2} = x\cos(2x)dx \Rightarrow \int\frac{dy}{y^2} = \intx\cos(2x)dx$$
$$u = x$$
$$\frac{du}{dx} = 1$$
$$\frac{dv}{dx} = \cos2x$$
$$v = \frac{1}{2}\sin(2x)}$$
$$\Rightarrow \frac{xsinx}{2} - \frac{1}{2}\int\sin(2x) = \frac{x\sinx}{2} + \frac{1}{4}\cos(2x) + C$$

So we have

$$\frac{-1}{y} = \frac{x\sin(x)}{2} + \frac{1}{4}\cos(2x) + C$$

So

$$y = \frac{-4}{\cos(2x)} - \frac{2}{x\sin(2x)} + C$$

$$\mbox{at y(0) = -1 \Rightarrow 1 = \frac{-4}{1} - \frac{2}{0} + C \Rightarrow C = 5}$$

So overall,

$$y = \frac{-2}{x\sin(2x)} - \frac{4}{\cos(2x)} + 5}.$$

Last edited by a moderator: Feb 8, 2006
2. Feb 8, 2006

Tom Mattson

Staff Emeritus
OK.

Nope. You can't take the reciprocal of a fraction that way. If you could then we would have the following:

$\frac{1}{2}=\frac{1}{3}+\frac{1}{6}$ (True)

So

$2=3+6$ (False)

Instead you must combine the two terms on the right with a common denominator, and then take the reciprocal.

3. Feb 8, 2006

jamesbob

$$\mbox{2. Find a particular integral for each of the following equations:}$$

$$/mbox{ i. \frac{d\theta}{dz} + 2\theta = 8}$$
$$/mbox{ ii. \frac{dx}{dt} - 2x = 14e^{-5t}}$$
$$/mbox{iii. \frac{dx}{dt} + x - -3sin2t + 4cos2t}$$

$$\mbox{ i. Constant term so choose x = a + bt}$$
$$\theta(z) = 1 \Rightarrow \frac{d\theta}{dz} = 0 \Rightarrow 0 + 2a = 8 \rightarrow a = 4 \Rightarrow PI = \Theta(z) = 4.$$

$$\mbox{ ii. Choose x = ae^(-t). This gives -5ae^{-5t} - 2ae^{-5t} = 14e^{-5t} \Rightarrow a = -2 \Rightarrow PI: x(t) = -2e^{-5t)}$$

$$\mbox{ iii. All i know to do here so far is choose x = acos2t + bsin2t. How do i continue?}$$

4. Feb 8, 2006

jamesbob

Sorry il sort the coding to this when i have a second

5. Feb 13, 2006

jamesbob

$$\mbox{2. Find a particular integral for each of the following equations:}$$

$$i. \frac{d\theta}{dz} + 2\theta = 8$$
$$ii. \frac{dx}{dt} - 2x = 14e^{-5t}$$
$$iii. \frac{dx}{dt} + x - -3sin2t + 4cos2t$$

$$i. Constant term so choose x = a + bt$$
$$\theta(z) = 1 \Rightarrow \frac{d\theta}{dz} = 0 \Rightarrow 0 + 2a = 8 \rightarrow a = 4 \Rightarrow PI = \Theta(z) = 4.$$
$$ii. Choose x = ae^(-t). This gives -5ae^{-5t} - 2ae^{-5t} = 14e^{-5t} \Rightarrow a = -2 \Rightarrow PI: x(t) = -2e^{-5t)$$