Question

) For what values of k does the function y = cos(kt) satisfy the differential equation 4y 00 = −25y? (1) (b) For those values of k that you found in part (a), verify that every function of the form y = A sin(kt) + B cos(kt) is also a solution to the differential equation (1). (Here, A and B are constants.)

Answer 1

Answer:

a.$k=\pm\frac{5}{2}$

Step-by-step explanation:

We are given that a solution

y= cos (kt) satisfied the differential equation

$4y''=-25y$

We have to find  the value of k

a.y=coskt

Differentiate w.r.t x

Then we get

$y'=-k sinkt$

$\frac{d(cos ax)}{dx}=-asin ax$

Again differentiate w.r.t x

$y''=-k^2 cos kt$ ($\frac{d(sinax)}{dx}=a cos ax$

Substitute the value in given differential equation

$-4 k^2 coskt=-25 coskt$

coskt cancel on both sides then we get

$4k^2=25$

$k^2=\frac{25}{4}$

a.$k=\sqrt{\frac{25}{4}}=\pm\frac{5}{2}$

b.We have to show that y=A sin kt + B cos kt is a solution to given differential equation for k=$\pm\frac{5}{2}$

Substitute the values of k

Then we get

$y=A cos \frac{5}{2} kt+ B sin \frac{5}{2} kt$

Differentiate w.r.t x

$y'=-\frac{5}{2} A sin \frac{5}{2}t+ \frac{5}{2} B cos \frac{5}{2}t$

Again differentiate w.r.t x

Then we get

$y''=-\frac{25}{4}A cos \frac{5}{2} t+\frac{25}{4} B sin\frac{5}{2} t$

Substitute the value of y'' and y in given differential equation

$-25 A cos \frac{5}{2} t +25 B sin \frac{5}{2} t=-25(A cos \frac{5}{2} t+ B sin \frac{5}{2} t)$

$-25 A cos \frac{5}{2} t +25 B sin \frac{5}{2} t=-25 y$

LHS=RHS

Hence, every function of the form

y=A cos kt +B sin kt is a s solution of given differential equation for k=$\pm\frac{5}{2}$

Where A and B are constants