Show that any function of the form : y = A sinh mx + B cosh mx satisfies the differential equation y^n = ym^2.

Question

(a) Show that any function of the form : y = A sinh mx + B cosh mx  satisfies the differential equation yⁿ = m²y.
(b) Find y = y(x) such that yⁿ = 9y, y(0) = -4, and y'(0) = 6.

 

I don't even know where to start

 

#55, p263

ANSWER: y = 2 sinh (3x) - 4 cosh (3x)

Answer 1

Hi Mia,

 

I infer from the answer to part b that you have stated the problem incorrectly. Looks to me like where you have yⁿ in ^{the problem statements, that in fact it should be} y'' ^{(y double prime, i.e. second derivative).}

 

Proceeding under that assumption and using the fact that D_xsinh u = cosh u D_xu and D_xcosh = sinh u D_xu, we have:

 

y^' = Acosh mx (m) + B sinh mx (m) = m (A cosh mx + B sinh mx)

y^''= m(Asinh mx (m) + B cosh mx (m)) = m²(Asinh mx + B coshmx); and since the expression in () is equal to y we have y^''=m²y (Q.E.D.)

 

For part b) for now, consider starting with the answer, taking the 1st and 2nd derivatives and verify that the conditions of the problem are indeed satisfied. That might give a clue how to solve using anti-derivatives.

 

Update for b)

 

Since y^''=9y, we have y^''= 9(Asinh mx + Bcosh mx).

 

We should be able to get y' by taking anti-derivative on both sides, inferring the constant of integration is 0.

 

y^'= (9A/m)cosh mx + (9B/m) sinh mx

 

and

 

y = (9A/m²)sinh mx + (9B/m²)cosh mx

 

At this point we have three unknowns, A, B, m. 

 

Try using y(0) = -4, y'(0) = 6 to deduce those values.

 

I see equations like:

 

y(0) = 9A/m² = -4

y = (9/m²)y

y^'(0) = 9A/m = 6

 

Trying to arrive at the conclusion that A = 2, B = -4, m = 3.