How to solve the two following differential equations?

In summary, when the second derivative of a function vanishes, it can at most be linear in x. Similarly, when the second derivative of a function is equal to k^2 times the function, the solution can be expressed as Ae^(-kx) + Be^(-kx). The solutions can be deduced by integrating both sides of the equations and considering different cases, such as when k^2 is positive, zero, or negative. Another approach is using a constant coefficient linear differential equation and assuming a solution of the form e^(rx), and then solving for the inverse.
  • #1
Tspirit
50
6
(1) ##\frac{d^{2}y}{dx^{2}}=0##
(2) ##\frac{d^{2}y}{dx^{2}}=k^{2}y##, where k is a real positive number.
 
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  • #2
If the second derivative of a function vanishes, it can at most be linear in x. Same reasoning for your second equation, although looking at the notation I suspect you mean the rhs to have y(x) instead of x. If that is so, you want to try an exponential function like e^[kx].
 
  • #3
haushofer said:
If the second derivative of a function vanishes, it can at most be linear in x. Same reasoning for your second equation, although looking at the notation I suspect you mean the rhs to have y(x) instead of x. If that is so, you want to try an exponential function like e^[kx].
Thank you for your answer. But for the first equation, is there any process of deduction?
 
  • #4
haushofer said:
If the second derivative of a function vanishes, it can at most be linear in x. Same reasoning for your second equation, although looking at the notation I suspect you mean the rhs to have y(x) instead of x. If that is so, you want to try an exponential function like e^[kx].
Thank you very much for pointing the spelling error in the second equation, I have modified it.
 
  • #5
haushofer said:
If the second derivative of a function vanishes, it can at most be linear in x. Same reasoning for your second equation, although looking at the notation I suspect you mean the rhs to have y(x) instead of x. If that is so, you want to try an exponential function like e^[kx].
I have searched google for "harmonic oscillator", however, it is a little different to the equation I give. For harmonic oscillator, the equation is like this, ##\frac{d^{2}y}{dx^{2}}=-k^{2}y##, where there is an additional negative sign.
 
  • #6
Now I know the solutions:
(1)For $$\frac{d^{2}y}{dx^{2}}=0,$$ the solution is $$y=A+Bx.$$
(2)For $$\frac{d^{2}y}{dx^{2}}=k^{2}y,$$ the solution is $$y=Ae^{-kx}+Be^{-kx}.$$
But I still don't know how they are deduced.
 
  • #7
Tspirit said:
Now I know the solutions:
(1)For $$\frac{d^{2}y}{dx^{2}}=0,$$ the solution is $$y=A+Bx.$$
(2)For $$\frac{d^{2}y}{dx^{2}}=k^{2}y,$$ the solution is $$y=Ae^{-kx}+Be^{-kx}.$$
But I still don't know how they are deduced.
1) is easy. Just integrate both sides twice, remembering to plug in a constant of integration.
For 2), multiply both sides by dy/dx. Now you can integrate both sides. What do you get?
 
  • #8
Tspirit said:
(2)For $$\frac{d^{2}y}{dx^{2}}=k^{2}y,$$ the solution is $$y=Ae^{-kx}+Be^{-kx}.$$
But I still don't know how they are deduced.
That is a constant coefficient linear DE of the form ##y'' + \lambda y = 0##. The standard technique for such an equation is to assume a solution of the form ##e^{rx}##. Plug that in and see what you get. You will want to consider three cases:$$
\lambda = k^2 > 0,~\lambda = 0,~\lambda = -k^2 <0\text{.}$$
 
  • #9
LCKurtz said:
The standard technique for such an equation is to assume a solution of the form
Sure, but I interpreted Tspirits' question as how to proceed if you have no idea what the answer is.
 
  • #10
$$\dfrac{d^2y}{{dx}^2}-k^2y=0$$
we have haruspex's sugestion
$$2\dfrac{d^2y}{{dx}^2}\dfrac{dy}{dx}-2k^2y\dfrac{dy}{dx}=0 \\
\dfrac{d}{dx}\left(\dfrac{dy}{dx}+k y\right)\left(\dfrac{dy}{dx}-k y\right)=0$$
we could factor the operator
$$\left(\dfrac{d}{dx}+k \right)\left(\dfrac{d}{dx}-k \right)y=0$$
we could solve for the inverse
$$-\left(\dfrac{dx}{dy}\right)^{-3}\dfrac{d^2x}{{dy}^2}-k^2y=0$$
 
  • #11
lurflurf said:
haruspex's sugestion
I was thinking more along the lines of
y"=k2y
2y'y"=2k2yy'
y'2=k2y2+c2
Then some suitable hyperbolic trig substitution.
 

FAQ: How to solve the two following differential equations?

1. What is a differential equation?

A differential equation is a mathematical equation that relates a function with its derivatives. It describes how a quantity changes over time or space, and is used to model many physical, biological, and economic phenomena.

2. What are the two types of differential equations?

The two types of differential equations are ordinary differential equations (ODEs) and partial differential equations (PDEs). ODEs involve a single independent variable, while PDEs involve multiple independent variables.

3. How do you solve a differential equation?

The method for solving a differential equation depends on its type and complexity. Some common methods include separation of variables, integrating factors, and using numerical methods such as Euler's method or Runge-Kutta methods.

4. What is the importance of solving differential equations?

Differential equations are important in many fields of science and engineering as they allow us to model and understand complex systems and phenomena. They are used in areas such as physics, biology, economics, and engineering to make predictions and solve problems.

5. What are some real-life applications of differential equations?

Differential equations have numerous applications in real life, such as in modeling population growth, predicting the spread of diseases, describing the behavior of electrical circuits, and analyzing the motion of objects under the influence of forces. They are also used in engineering to design and optimize systems such as bridges, airplanes, and cars.

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