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## Differential equations

### Course: Differential equations>Unit 2

Lesson 1: Linear homogeneous equations

# 2nd order linear homogeneous differential equations 3

Let's use some initial conditions to solve for the particular solution. Created by Sal Khan.

## Want to join the conversation?

• What if the characteristic equation turns out to be quadratic? for instance (r^2 - 6r + 9).
The factors of that would then be (r-3)^2. Does your example here infer that the general solution of that would be: y(x) = c1e^(3x) + c2e^(3x)? (and not just ce^(3x)) •  Excellent question.

If the characteristic equation turns out to be a quadratic there are three possible scenarios. In this video only one scenario is explored, namely, when the roots of the characteristic equation are real and different. In this case the general solution will always work.

The second scenario is when the roots are real but identical (like the example you gave). Notice that your solution can be rewritten by factoring out the like term e^(3x) giving you, y(x) = (c1+c2)*e^(3x)
And since a constant plus a constant is a constant,
y(x)=c*e^(3x). However, this is not a sufficient general solution because it only has one part (this is a terribly oversimplified explanation). There is a general solution for this scenario but it is different. Try searching for it.

The third scenario is when the roots are not real, i.e. they are imaginary. For example, try solving the equation y''+ y '+ 1 = 0
Thumbs up!
• what's the method to find a solution to this 2nd order linear homogeneous differential equations. we can't just guess and try a solution. in this example you suggested e to the rx and it worked but there must be a way ti find that a solution. Thanks. • The "method" Sal used here was to "choose" a function whose derivatives were multiples of itself as a possible solution the differential equation. As the only way to sum multiples of a function and its derivatives such that they equal zero is for the function and its derivatives to be multiples of one another. Well, we've already learned that the only function that satisfies this condition is f(x)= e^(rx) since d^n/dx^n(f(x))=r^n*e^(rx), i.e., the nth derivative of e^(rx) is just r^n*e^(rx) which itself is a multiple of the original function. Thus, f(x)=e^(rx) is a general solution to any 2nd order linear homogeneous differential equation.
To find the solution to a particular 2nd order linear homogeneous DEQ, we can plug in this general solution to the equation at hand to find the values of r that satisfy the given DEQ. These particular values of r give general solutions which themselves can be combined linearly to form a more general solution to the original DEQ. To get the specific solution to the DEQ, one must use initial conditions to calculate the coefficients that apply to the solution for this particular DEQ.
This "guess & check" method is employed all the time in advanced science and engineering courses. Your intuition about the problem leads to an answer which you then verify.
The neat thing about this method for the solution of homogeneous 2nd order DEQs is that the solution boils down to simple algebra. The characteristic equation derived by differentiating f(x)=e^(rx) is a quadratic equation for which we have several methods to easily solve.
Furthermore, if the solutions to the characteristic equation are real, we get solutions that involve exponential growth/decay. However, if the solutions of the characteristic equation are imaginary, we get trigonometric/periodic functions (sin x & cos x) as solutions to the DEQ since e^(irx)= cos(rx)+ isin(rx). Hence, f(x)=e^(rx) is a very versatile function.
• Can you show this using a matix? • how can i determine if an equation is linear or non linear equation.... can you give me some examples • Angelique,

In Linear DE's, the variable that is being differentiated isn't itself part of a non-linear function.

For example, in the differential equation y'' +3y' +y=7x+2, the variable that is being differentiated is y. This differential equation is linear, because there are no y^2, y^3, e^y, cos (y), sin ( y' ) , yy' terms, or anything like that.

Note, it would be okay to multiply a y term (or y', y'', y''', etc) by any constant, and the equation would still be linear. In the above equation, for example, there is a product of the 1st derivative of y and 3... that doesn't make the equation non-linear. In fact, it's even ok to multiply the term being differentiated by the independent variable and the equation is still linear: y''+xy'+y=7x is still linear... and so is y'' +sin(x) y' + y=7x.

But all of the following equations are NON-linear:

ln( y'') = y'
y''/y = y' (But y''/y=4x is linear, because you can manipulate it to: y'' -4xy =0)
sin (y')=4x
(y'')^2+y =0

et cetera, et cetera.... make sense?
• what does the solution of a differential equation actually mean? • What does the solution to a regular equation mean? The solution set of 3x + 1 = 10 is the set of x's that satisfy 3x + 1 = 10. One solution is one element in that solution set. For differential equations, it's the same, except the solutions are functions. The solution set of f'(x) + f(x) = x^2 is all functions f that satisfy that equation (for all x).
• How do I deal with a situation in which I have an equation that contains both a function, it's derivatives and the variable that the function is defined in terms of?

For instance, if I have 20y'' + y' = 150 - 2x
Where y is defined in terms of x. • hello, is it possible to find c1 and c2 using non-graphing calculator? • Find a second solution of the differential equation
x^2y''+-xy'+2y=0 and y=xsin(lnx)
by using the formula.
how can we solve this.?   