Question

Asked Feb 19, 2020

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Find y as a function of t if

5y′′+31y=0

y(0)=4,y′(0)=7

y(t)=

Solving the given differential equation

The given differential equation is

5y''+31y=0 ......(1)

and the initial condition is y(0)=4 and y'(0)=7

To get the auxiliary equation we replace y by 1 and y'' by m^{2}

So, the auxiliary equation is

5m^{2}+31=0

The roots of the above equation are

so, the solution of the differential equation is

......(1)

where A and B are arbitrary constants

Find the arbitrary constants

Using y(0)=4 we get

4=A*1+B*0

A=4

Differentiating (1) with respect to t we get

Now using y'(0)=7 we get

...

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