Author Topic: differential equations (Read 1364 times) Tweet Share

Chavi · « **on:** June 18, 2010, 01:52:25 pm »

Hey, can I please get some help with this question?

9. Find the equation of the curve whose rate of change of gradient with respect to x at any point is proportional to the y coordinate of the point, given that when $x=0, y=1, \frac{dy}{dx} = 2, and, \frac{d^2y}{dx^2} = 4$ Thanks

kenhung123 · « **Reply #1 on:** June 18, 2010, 02:47:12 pm »

huh whats d^2 x^2?

brightsky · « **Reply #2 on:** June 18, 2010, 03:13:48 pm »

Quote from: kenhung123 on June 18, 2010, 02:47:12 pm

huh whats d^2 x^2?

Double derivatives.

/0 · « **Reply #3 on:** June 18, 2010, 03:15:45 pm »

$\frac{d^2y}{dx^2}$ is the second derivative, or $\frac{d}{dx}\left(\frac{dy}{dx}\right)$

For the question, the wordy bit is saying that

$\frac{d^2y}{dx^2} \propto y$ , or $\frac{d^2y}{dx^2} = ky$ , where $k$ is a constant.

Then solve and use the given information to find $y$ in terms of $x$ .

Chavi · « **Reply #4 on:** June 18, 2010, 03:42:48 pm »

Quote from: /0 on June 18, 2010, 03:15:45 pm

$\frac{d^2y}{dx^2}$ is the second derivative, or $\frac{d}{dx}\left(\frac{dy}{dx}\right)$

For the question, the wordy bit is saying that

$\frac{d^2y}{dx^2} \propto y$ , or $\frac{d^2y}{dx^2} = ky$ , where $k$ is a constant.

Then solve and use the given information to find $y$ in terms of $x$ .

I tried that. Can't seem to get the answer, particularly trying to integrate the expression for dy/dx

/0 · « **Reply #5 on:** June 18, 2010, 06:24:29 pm »

$\frac{d^2y}{dx^2} = ky$

In Essentials there is a useful kinematics formula which can be applied here: $\frac{dv}{dt} =\frac{d(\frac{1}{2}v^2)}{dx}$ . Where $v = \frac{dx}{dt}$ .

Expressed another way, it is $\frac{d^2x}{dt^2} = \frac{d}{dx}\left(\frac{1}{2}\left(\frac{dx}{dt}\right)^2\right)$

If we now make the change $x \to y$ , $t \to x$ , then we get:

$\frac{d^2y}{dx^2} = \frac{d}{dy}\left(\frac{1}{2}\left(\frac{dy}{dx}\right)^2\right)$

Hope that helps

Chavi · « **Reply #6 on:** June 19, 2010, 07:16:53 pm »

Still not getting it. .
How do you integrate the derivative that turns out to k/(ln(y) + 1/4 = dy/dx ?

/0 · « **Reply #7 on:** June 20, 2010, 12:47:27 am »

$\frac{1}{2}\left(\frac{dy}{dx}\right)^2 = \frac{1}{2}ky^2+C$

$\left(\frac{dy}{dx}\right)^2 = ky^2+D$

$\frac{dx}{dy} = \frac{1}{\sqrt{ky^2+D}}$

hm... i don't think this kind of question would be in the scope of specialist maths.

If you want to look for the general method of solving differential equations like this, google
"Second Order Linear Homogenous Constant Coefficient Differential Equations"

mark_alec · « **Reply #8 on:** June 20, 2010, 12:58:43 am »

Isn't the answer our favourite oscillating answer?

$y = Ae^{\pm\sqrt{k}x}+Be^{\pm\sqrt{k}x}$

Mao · « **Reply #9 on:** June 20, 2010, 04:46:52 am »

@marc_alec, very sneaky question. This is a 2nd order ODE in the second order, but /0's method is within the scope and workable if we substitute everything to get rid of the constants.

@Chavi, highly unlikely that you'll have to do this in an exam [in fact, I'd say impossible]. You wouldn't really have apply a kinematics thing to this kind of 'general' case [unless they give you hints]. I will finish the rest of the question from where /0 has left it.

Quote from: /0 on June 18, 2010, 06:24:29 pm

$\frac{d^2y}{dx^2} = ky$ ------[1]
$\frac{d^2y}{dx^2} = \frac{d}{dy}\left(\frac{1}{2}\left(\frac{dy}{dx}\right)^2\right)$ --------[2]

Given that when x=0, y = 1 and d²y/dx² = 4, we then have (by [1]) that 4 = k * 1, which implies k=4.

After integration, we have

Quote from: /0 on June 20, 2010, 12:47:27 am

$\left(\frac{dy}{dx}\right)^2 = ky^2+D$ ----------[3]

Given that when x=0, y = 1, dy/dx = 2, thus by [3], 2^2 = 4 * 1^2 + D, which implies D = 0, [3] can thus be simplified to

$\left(\frac{dy}{dx}\right)^2 = 4y^2$

$\implies \frac{dy}{dx} = 2 y$ , taking only the positive branch of the square root since both y and dy/dx are positive.

$\implies \frac{dx}{dy} = \frac{1}{2y} \implies x = \frac{1}{2}\log_e |y| + C$ , where C = 0 by substituting x=0, y=1.

$\implies |y| = e^{2x}$

$\implies y = e^{2x}$ , taking only the positive branch since y is positive.

Chavi · « **Reply #10 on:** June 20, 2010, 05:05:30 pm »

thanks mao

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