How to Determine a Differential Equation from a Given Solution?

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To determine a differential equation from the solution y = C1sin(3x) + C2cos(3x), it is essential to recognize that the original equation must have complex roots of 3i and -3i. The characteristic equation is λ² + 9 = 0, leading to the ordinary differential equation (ODE) d²y/dx² + 9y = 0. Differentiating the given solution twice helps eliminate the constants C1 and C2, facilitating the formation of the ODE. The process involves recognizing the form of the solution and applying differentiation to derive the necessary relationships. Ultimately, this method highlights the connection between the solution's structure and the corresponding differential equation.
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Q. Determine a homogeneous linear differential equation with constant coefficients having having the following solution:

y = C1sin3x + C2cos3x

My idea is to differntiate both sides with respect to x and come up with an equation in dy/dx

what else? can be done...

Is my idea correct.
 
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Hello,

Since the solution is in the form y=ASin3x+BCos3x,
The original equation must have complex roots 3i and -3i.
Thus, a possible solution is d^2y/dx^2+9=0. =)
 
Not every differential equation is a first order equation!
 
estalniath said:
Hello,

Since the solution is in the form y=ASin3x+BCos3x,
The original equation must have complex roots 3i and -3i.
Thus, a possible solution is d^2y/dx^2+9=0. =)

Pay attention.The characteristic equation is

\lambda^{2}+9=0

,but the ODE is

\frac{d^{2}y}{dx^2}+9y=0

Okay?


Daniel.
 
Can somebody explain how they arrived at \lambda^{2}+9=0

I know that the two roots are 3i and -3i. I had figured out this already.
 
Suppose you were given \lambda^{2}+9=0

How would factor it , in order to find the two values for \lambda
 
By multiplying (\lambda-3i)(\lambda+3i) and equating it to 0...?

Daniel.
 
Yup I got it thanks!
 
Thanks for pointing that out Daniel! I guess that I took the "y" there for granted every time I used the characteristic solution to get the y_h
 
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By the way- this was clearly a simple problem because the given combination was clearly a solution to a linear equation with constant coefficients. It's not always that simple. In general, given a combination of functions with TWO "unknown constants", you form the simplest equation, involving differentials, the eliminates those constants.

If you did NOT recognize y= C1cos(3x)+ C2sin(3x) as coming from λ= 3i and -3i, you could have done this:
Since you are seeking a differential equation: DIFFERENTIATE-
y'= -3 C1 sin(3x)+ 3 C2 cos(3x).
Since there are two unknown constants, DIFFERENTIATE AGAIN-
y"= -9 C1 cos(3x)- 9 C2 sin(3x).

Now do whatever algebraic manipulations you need to eliminate the two constants.

(In this example, of course, just add y" and 3y.)
 

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