Verify the statement of Example 2 . Also verify that y=coshx and y=sinhx are solutions of y^''=y

Name: Problem 1, Chapter 8: Mathematical Methods in the Physical Sciences
Uploaded: 2022-01-05T12:27:08-08:00
Duration: 4 min 43 s
Channel: Aditya Jain
Description: Verify the statement of Example 2 . Also verify that y=coshx and y=sinhx are solutions of y^''=y

Verify the statement of Example 2 . Also verify that y=coshx...

Key Concepts

Verification Process in Differential Equations

The verification process involves differentiating the proposed solution, computing its derivatives up to the required order, and then substituting these into the differential equation. If the left-hand side simplifies to the right-hand side for all x in the domain, the function is a valid solution. This method is commonly used to confirm that functions like cosh x and sinh x satisfy equations such as y'' = y.

Hyperbolic Functions

Hyperbolic functions, such as cosh x and sinh x, are analogues of the trigonometric functions but for hyperbolic geometry. They are defined using exponential functions, where cosh x = (e^x + e^(-x))/2 and sinh x = (e^x - e^(-x))/2. Their derivatives have properties similar to trigonometric functions, with cosh' x = sinh x and sinh' x = cosh x, making them useful in solving differential equations.

Second Order Homogeneous Linear Differential Equations

These are differential equations of the form y'' + ay' + by = 0, where a and b are constants. They frequently appear in modeling physical phenomena such as vibrations and electrical circuits. The process of verifying solutions involves computing the derivatives of a proposed solution and substituting back into the equation to check if the equation holds true, affirming that the function is indeed a solution.

Transcript

00:01 All right, so i've placed the question here and it says verify the statement of example two.

00:06 Now before we move on to the next question, let's try and do this first.

00:11 So we have to verify example two.

00:12 And example two says, let me write down what example two says.

00:16 Example two says that the equation y double prime equals y has solutions.

00:30 So the solutions are y equals e.

00:36 To the x y equals e to the minus x the third one is a linear combination of both of these a to the x plus b e to the minus x so first we have to verify this and let's do that so let's press do it you know sequentially let's try and do the first one we have y equals e to the x and to verify this equation we just have to put the left hand side to be equal to our h our right hand side right and so let's see, we take the first derivative of this of y, which uses y prime, and the derivative of an exponential is just another exponential, multiplied the, by the derivative of the argument, but the argument is just x, so it's going to be just one.

01:21 Let me hide this.

01:22 And the second derivative is also going to be e to the x.

01:27 And so clearly what we can find, what we can see is y double prime is the same as y, and it's the same as e.

01:34 To the x okay let's try and do the second one we have y equals e to the minus x now the first derivative is going to look like a to the minus x the derivative of an exponential is just the same except now we'll have to multiply this by the derivative of the argument which is going to be minus one so we'll get a minus in front of this and the second derivative is also going to look like minus e to the minus x but now again we'll multiply with the derivative of the argument and we'll get a plus and so we can just if effectively erase this thing.

02:07 And clearly we can see that again, we find that the y and the y prime prime or y double prime are the same.

02:17 Okay.

02:17 So we have verified for the first two.

02:20 Let's look at this third one, which again an easy extension of both of these.

02:24 We have a linear combination.

02:26 This is also called a linear combination...

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