206k views
4 votes
Suppose that a fourth order differential equation has a solution y=-7 e^{2 x} x sin (x) \). iai Find such a differential equation, assuming it is homogeneous and has constant coefficie

1 Answer

4 votes

To find a differential equation with a given solution, we can differentiate the given solution multiple times and substitute it into a general form of a differential equation.
Let's start by finding the first derivative of the given solution:
y = -7e^(2x) * x * sin(x)
To differentiate this expression, we can use the product rule:
y' = (-7e^(2x) * x * cos(x)) + (-7e^(2x) * sin(x)) + (-7e^(2x) * x * sin(x))
Simplifying this expression gives:
y' = -7e^(2x) * (x * cos(x) + sin(x) + x * sin(x))
Next, let's find the second derivative:
y'' = -7e^(2x) * [(cos(x) - x * sin(x)) + (cos(x) + x * cos(x) + sin(x))]
Simplifying further:
y'' = -7e^(2x) * [2cos(x) + x * (cos(x) - sin(x))]
Continuing this process, we can find the third and fourth derivatives:
y''' = -7e^(2x) * [2cos(x) + 3(cos(x) - sin(x)) + x * (-sin(x) - cos(x))]
y'''' = -7e^(2x) * [2cos(x) + 6(cos(x) - sin(x)) + x * (-3sin(x) + 2cos(x))]
Now, let's substitute the original solution and its derivatives into the general form of a fourth-order homogeneous differential equation with constant coefficients:
ay'''' + by''' + cy'' + dy' + ey = 0
Substituting the derivatives into the equation, we get:
a[-7e^(2x) * [2cos(x) + 6(cos(x) - sin(x)) + x * (-3sin(x) + 2cos(x))]]
+ b[-7e^(2x) * [2cos(x) + 3(cos(x) - sin(x)) + x * (-sin(x) - cos(x))]]
+ c[-7e^(2x) * [2cos(x) + x * (cos(x) - sin(x))]]
+ d[-7e^(2x) * [x * cos(x) + sin(x) + x * sin(x)]]
+ e[-7e^(2x) * x * sin(x)] = 0
Simplifying this equation will give us the desired fourth-order homogeneous differential equation with constant coefficients.

User Batu
by
7.9k points

Related questions

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories