# Verify y1 solves 2nd order differential equation

• accountkiller
In summary, the conversation discusses verifying if the function y1 = (e^x) * (cos x) satisfies the linear, homogeneous, 2nd order differential equation y'' - 2y' + 2y = 0. The individual used the product rule and integration by parts, but after some confusion, it was discovered that only differentiation and substitution were needed to show that y1 is a solution to the equation.
accountkiller

## Homework Statement

Verify that y1 = (e^x) * (cos x) solves the linear, homogeneous, 2nd order differential equation y'' - 2y' + 2y = 0

## The Attempt at a Solution

So I used the product rule but just kept going in circles with the cos x. I remember there's a way to do this so you don't go in circles, but I forgot! I'd appreciate any help.

What did you get when you differentiated it?

So the product rule is fg' = fg - int(gf').
I said f = cos x, so f' = -sin x dx, and g' = e^x, so g = e^x.
This gives me e^x(cos x) - int(e^x(-sin x) dx) = e^x(cos x) + int(e^x(sin x) dx).
So then I have to use the product rule again for that integral, with f = e^x, f' = e^x dx, g' = sin x, and g = -cos x. This gives me -e^x(cos x) + int(e^x(cos x) dx) for that integral, and putting that together with what was in front of it:
y1' = e^x(cos x) - e^x(cos x) - int(e^x(cos x) dx)
y1' = -int (e^x(cos x) dx)

So I'm pretty much back where I started.

You do not need to integrate anything.

Just find y1'', y1' and you have y, put back into the DE and see if you get zero. If you do, it satisfies the DE.

OH! I was using integration by parts instead of the product rule. Oi, I shouldn't get integration and differentiation mixed up :P

Thanks!

## 1. What is a 2nd order differential equation?

A 2nd order differential equation is a mathematical equation that involves the second derivative of an unknown function. It is often used to model physical systems and describe the relationship between a function and its rate of change.

## 2. How do you verify that y1 solves a 2nd order differential equation?

To verify that y1 solves a 2nd order differential equation, we can substitute y1 into the equation and check if the resulting equation is satisfied. This means that the second derivative of y1 should equal the right-hand side of the differential equation.

## 3. What are the steps to verify a solution to a 2nd order differential equation?

The steps to verify a solution to a 2nd order differential equation are:
1. Substitute the proposed solution into the equation.
2. Take the second derivative of the solution.
3. Simplify the resulting equation.
4. Check if the simplified equation is equal to the right-hand side of the original differential equation. If it is, then the proposed solution is verified.

## 4. Can a 2nd order differential equation have multiple solutions?

Yes, a 2nd order differential equation can have multiple solutions. This is because the general solution to a 2nd order differential equation contains two arbitrary constants, which can take on different values and result in different solutions.

## 5. How is a 2nd order differential equation used in real life?

2nd order differential equations are used in various fields such as physics, engineering, and economics to model real-life systems and phenomena. For example, they can be used to describe the motion of a pendulum, the growth of a population, or the behavior of an electrical circuit.

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