MHB -1.8.17 initial value then find C

[karush]

$\tiny{1.8.17}$
\nmh{324]
Use direct substitution to verify that y(t) is a solution of the given differential equation in . Then use the initial conditions to determine the constants C or $c_1$ and $c_1$
$y''+4y=0, \quad y(0)=1,\quad y'(0)=0,\quad y(t)=c_1\cos 2t+c_2\sin 2t$

 

[cbarker1]

What are the directions asking you to do? What are the second derivatives of $\cos(2t)$ and $\sin(2t)$?

 

[karush]

well...
$y'(t)=-2c_1\sin (2t)+2c_2\cos (2t)$
$y''(t) =-4c_1\cos (2t)-4c_2\sin (2t)$

 

[HOI]

The problem said "use direct substitution"! So substitute those into the given equation.

 

[karush]

The problem said "use direct substitution"! So substitute those into the given equation.

$y''+4y=0$
so then
$-4c_1\cos (2t)-4c_2\sin (2t)+4(c_1\cos 2t+c_2\sin 2t)=0$

ok by observation it all cancels out to 0
$y(0)=1$
$y(0)=c_1\cos 0+c_2\sin 0=c_1(1)+c_2(0)=1$

$y'(0)=0$
$y'(0)=-2c_1\sin (0)+2c_2\cos (0)=-2c_1(0)+2c_2(1)=0$

not sure is this is how you get the c values

 

[HOI]

Yes, that's the point.

Now, use the fact that $y(0)= c_1 cos(2(0))+ c_2 sin(2(0))= 1$
and $y'(0)= -2c_1 sin(2(0))+ 2c_2 cos(2(0))= 0$.

 

[karush]

i don't see that we are getting consistence values for $c_1$ and $c_2$

 

[HOI]

Since cos(0)= 1 and sin(0)= 0, those two equations are
$c_1= 1$ and $c_2= 0$,

What's wrong with that?

 

[karush]

but is that true for both y(0)=1 and y'(0)=0

 

[HOI]

In post #5 YOU wrote

y(0)=c1cos0+c2sin0=c1(1)+c2(0)=1y(0)=c1cos⁡0+c2sin⁡0=c1(1)+c2(0)=1

So c1= 1 and

y′(0)=−2c1sin(0)+2c2cos(0)=−2c1(0)+2c2(1)=0y′(0)=−2c1sin⁡(0)+2c2cos⁡(0)=−2c1(0)+2c2(1)=0

so 2c2= 0.

 

[cbarker1]

I suggest that you substitute for $c1$ to be 1. Then, plugin the function into the differential equation to see if the function with the calculated value still gives you the right answer. Btw, it is ok for the initial values to have different values. Thus, it does not need to be consistence.

 

[karush]

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