Question

(a) Show that any function of the form \(y = Asinhmx + Bcoshmx\) satisfies the differential equation \(y'' = {m^2}y\).

(b) Find \(y = y\left( x \right)\) such that \(y'' = 9y\), \(y\left( 0 \right) = - 4\) and \(y'\left( 0 \right) = 6\).

Short answer

(a) It is proved that the function of the form \(y = A{\rm{sinh}}mx + B{\rm{cosh}}mx\) satisfies the differential equation \(y'' = {m^2}y\).

(b) \(y = 2\sinh 3x - 4\cosh 3x\)

Step-by-step solution

To show that any function \(y = Asinhmx + Bcoshmx\) satisfies the differential equation \(y'' = {m^2}y\)

(a) Consider the function \(y = A\sinh mx + B\cosh mx\).

Differentiate the function twice.

\(\begin{aligned}y' & = \frac{d}{{dx}}\left( {A\sinh mx + B\cosh mx} \right)\\y' & = mA\cosh mx + mB\sinh mx\\y'' & = \frac{d}{{dx}}\left( {mA\cosh mx + mB\sinh mx} \right)\\ & = {m^2}A\sinh mx + {m^2}B\cosh mx\\ & = {m^2}\left( {A\sinh mx + B\cosh mx} \right)\\ & = {m^2}y\end{aligned}\)

Hence, it is proved that the function of the form \(y = A{\rm{sinh}}mx + B{\rm{cosh}}mx\) satisfies the differential equation \(y'' = {m^2}y\)

Find \(y = y\left( x \right)\)

(b) From part (a) the solution of the function \(y = A\sinh 3x + B\cosh 3x\) is \(y'' = 9y\) with \(y\left( 0 \right) = - 4\) and \(y'\left( 0 \right) = 6\) then we have,

\(\begin{aligned}y\left( 0 \right) & = A\sinh \left( 0 \right) + B\cosh \left( 0 \right)\\ - 4 & = B\end{aligned}\)

And

\(\begin{aligned}y'\left( x \right) & = 3A\cosh 3x - 12\sinh 3x\\y'\left( 0 \right) & = 3A\cosh 3\left( 0 \right) - 12\sinh 3\left( 0 \right)\\6 & = 3A\\A & = 2\end{aligned}\)

Thus, put the above-obtained values of \(A\) and \(B\) in the function \(y = A\sinh 3x + B\cosh 3x\).

\(y = 2\sinh 3x - 4\cosh 3x\)

The required function is \(y = 2\sinh 3x - 4\cosh 3x\).