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Determine whether the given function is a solution to the given differential equation.

1. x = 2 cos t - 3 sin t, x" + x = 0

2. y = sin x + x^2, d^2y/dx^2 + y = x^2 + 2

3. x = cos 2t, dx/dt + tx = sin 2t

Respuesta :

oodaramola

Answer:

It is only 1 and 2 that is the solution to the differential equation

Step-by-step explanation:

1. x''=- 2 Cos t+ 3 sin t and x=2 cos t -3 sin t the addition give 0

2. d^2y/dx^2=- Sin x +2 and y =Sin x +x^2 the addition =x^2+2

but

3. dx/dt= -2 Sin 2t and tx =t Cos 2t =t( 2Cos t - 2 Sin t) the addition it can't equal Sin 2t

adioabiola

Among the three cases as in the question; the given function which are solutions to the given differential equation are; Case 1 and Case 2.

According to the question;

We are required to determine whether the given function is a solution to the given differential equation.

For Case 1:

  • x = 2 cos t - 3 sin t

  • x'' = -2 cos t + 3sin t

  • Therefore; x" + x = 0

For case 2:

  • y = sin x + x^2

  • d²y/dx² = - sin x + 2

  • Therefore, d²y/dx² + y = x² + 2

For case 3:

  • x = cos 2t

  • dx/dt = -2 Cos 2t

  • tx = t cos 2t.

  • dx/dt + tx = (t - 2) cos 2t. which is not equal to Sin 2t

Read more:

https://brainly.com/question/14410483

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