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Chapter 4: Q4E (page 157)

Verify that y=sin3t+2cos3tis a solution to the initial value problem

2y''+18y=0;y(0)=2,y'(0)=3

Find the maximum of|yt|for-鈭�<t<鈭�.

Short Answer

Expert verified

Herey=sin3t+2cos3t is a solution for the given differential equation, the maximum value y=2.236.

Step by step solution

01

Differentiate the value of y

To verify thaty=sin3t+2cos3tis a solution of 2y''+18y=0.

Find y'and y''.

role="math" localid="1664098240818" y'(t)=3cos3t-6sin3ty''(t)=-9sin3t-18cos3t

Substitute these two equations in the differential equation;

2(-9sin3t-18cos3t)+18(sin3t+2cos3t)=0-18sin3t-36cos3t+18sin3t+36cos3t=00=0

Therefore, LHS = RHS.

02

Verify boundary condition.

Now verify the boundary conditions;

y(0)=sin(3脳0)+2cos(3脳0)=2

And

y'(0)=3cos(3脳0)-6sin(3脳0)=3

Therefore, y=sin3t+2cos3tis a solution for the given differential equation.

Now plot the graph y=sin3t+2cos3tand from the graph, one can see the maximum value y=2.236.

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