1
$\begingroup$

I can now recognise the order and the type of differential equations.Let's say $\frac{dy}{dx} = x^2 - 1$ is a first order ODE, $\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right)^2 + y = 0$ is a second order ODE and so on. I am having trouble to obtain a differential equation from a given function. I could find the differential equation for $y = e^x(A \cos x + B \sin x)$ and the steps that I followed are as follows.

$ \begin{align*} \frac{dy}{dx} &= e^x(A \cos x + B \sin x) + e^x(-A \sin x + B \cos x) \\ &= y +e^x(-A \sin x + B \cos x) \tag{1} \\ \frac{d^2 y}{dx^2} &= \frac{dy}{dx} + e^x(-A \sin x + B \cos x) + e^x(-A \cos x - B \sin x) \\ &= \frac{dy}{dx}+\left(\frac{dy}{dx} - y\right) - y \end{align*} $ using the orginal function and $(1)$. Finally, $ \frac{d^2y}{dx^2} - 2 \frac{dy}{dx} + 2y = 0 ,$ which is the required differential equation.

Similarly, if the function is $y=(A\cos2t + B\sin2t)$, the differential equation that I get is $ \frac{d^2y}{dx^2} + 4y = 0 $ following similar steps as above.

My question is how do I obtain the differential equations for the following functions using similer procedures

$y = Ae^{3x} + Be^{2x}$

$xy = Ae^{x} + Be^{-x} + x^{2}$

I am not looking for a solution in determinant form using vector spaces or any other linear algebra/matrices.Please provide step by step solution for the function in order to obtain a particular differential equation.

Thank you in advance.

  • 0
    I do not understand the downvote. Can the downvoter at least explain the reasons for the vote, and possibly suggest improvements?2011-09-03

3 Answers 3

1

Eliminate $a$ and $b$ from $y = ae^{2x} +be^{3x}\tag{1}$

$\frac{dy}{dx} = 2ae^{2x} + 3be^{3x}\tag{2}$

$d^2y/dx^2 = 4ae^{2x} + 9be^{3x}\tag{3}$

$(1)\cdot 2 -(2)$ we get

$2y - \frac{dy}{dx} = -be^3x$

$b = e^{-3x} \left(\frac{dy}{dx} - 2y\right)\tag{4}$

$(2)*2 - (3)$

$2 \frac{dy}{dx} - \frac{d^2y}{dx^2} = - 3 b e^{3x}$

$b = 1/ 3e ^{3x} \left(\frac{d^2y}{dx^2} - 2 \frac{dy}{dx}\right)\tag{5}$

compare $(4)$ and $(5)$

$(dy/dx - 2y) = 1/3 \left(\frac{d^2y}{dx^2} - 2 \frac{dy}{dx}\right)$

$3\frac{dy}{dx} - 6y = \frac{d^2y}{dx^2} -2 \frac{dy}{dx}$

$\frac{d^2y}{dx^2} -5 \frac{dy}{dx} + 6y = 0$

  • 0
    Welcome to Math.SE! Please see whether I changed the meaning of anything, it was hard to predict the behaviour of some fractions. You can learn how to use $\LaTeX$ [here](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference).2013-12-26
3

An answer has already been given for $y=Ae^{3x}+Be^{2x}$, so we deal with $xy=Ae^x+Be^{-x}+x^2 \qquad\text{(Equation $1$)}$

Differentiate both sides of the given equation twice with respect to $x$. We first get $x\frac{dy}{dx}+y=Ae^x-Be^{-x}+2x,\qquad\text{(Equation $2$)}$ and then, differentiating again, $x\frac{d^2y}{dx^2}+2\frac{dy}{dx}=Ae^x+Be^{-x}+2 \qquad\text{(Equation $3$)}$

Now look at Equation $1$ and Equation $3$. The term $Ae^x+Be^{-x}$ occurs in each. Subtract, and it disappears. We obtain $x\frac{d^2y}{dx^2}+2\frac{dy}{dx}-xy=2-x^2.$

  • 1
    The third equation says $P=Q$. The first says $R=S$. From this it follows that $P-R=Q-S$. Or if you want to do it the long way, use the first equation to solve for $Ae^x+Be^{-x}$. You will get $Ae^x+Be^{-x}=xy-x^2$. Substitute $xy-x^2$ for $Ae^x+Be^{-x}$ in the third equation.2011-09-01
2

Generally, in such problems, if the function involves $k$ arbitrary constants, we expect to differentiate $k$ times. Then, we should eliminate the arbitrary constants and obtain a differential equation in terms of $ y, \frac{dy}{dx}, \frac{d^2y}{dx^2}, \ldots, \frac{d^ky}{dx^k}. $

We will see how this works out in the case of $ y = Ae^{3x} + Be^{2x} . \tag{1}$ Since there are two arbitrary constants $A$ and $B$, we expect to involve $y$ and its first two derivatives. Differentiating twice, we get $ \frac{dy}{dx} = 3A e^{3x} + 2B e^{2x}, \tag{2} $ and $ \frac{d^2 y}{dx^2} = 9 A e^{3x} + 4B e^{2x}. \tag{3} $

Now, how do we eliminate $A$ and $B$ from these equations? We can eliminate $B$ from $(2)$ and $(3)$ by subtracting a suitable multiple of $(1)$. That is, $(2) - 2 \times (1)$ gives:
$ \frac{dy}{dx} - 2y = A e^{3x}. \tag{4} $ Similarly, $(3) - 4 \times (1)$ gives: $ \frac{d^2y}{dx^2} - 4y = 5 A e^{3x}. \tag{5} $

To get the differential equation, we now need to similarly eliminate $A$ from these equations $(4)$ and $(5)$. Do you see how?

  • 0
    @alok Given the disagreement between us in the other post, I think it will be more comfortable for both us if I make this answer community wiki. That way, you can remain assured that I do not care only about my reputation points.2011-09-05