Solving Differential Equations with Constant Terms: Particular Integrals

[jamesbob]

Hey, just want to 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}

My answer:

\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}.

 

[quantumdude]

So we have

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

OK.

So

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

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[/color])

So

2=3+6 (False[/color])

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

 

[jamesbob]

\mbox{2. Find a <b>particular integral</b> 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}

My answers:

\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?}

 

[jamesbob]

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

 

[jamesbob]

That post shoul have read:

\mbox{2. Find a <b>particular integral</b> 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

My answers:

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)

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