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[SOLVED] 1.8.17 initial value then find C

karush

Well-known member
Jan 31, 2012
2,886
$\tiny{1.8.17}$
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 $c_1$
$y''+4y=0, \quad y(0)=1,\quad y'(0)=0,\quad y(t)=c_1\cos 2t+c_2\sin 2t$
 
Last edited:

Cbarker1

Active member
Jan 8, 2013
255
What are the directions asking you to do? What are the second derivatives of $\cos(2t)$ and $\sin(2t)$?
 

karush

Well-known member
Jan 31, 2012
2,886
well......
$y'(t)=-2c_1\sin (2t)+2c_2\cos (2t)$
$y''(t) =-4c_1\cos (2t)-4c_2\sin (2t)$
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
731
The problem said "use direct substitution"! So substitute those into the given equation.
 

karush

Well-known member
Jan 31, 2012
2,886
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
 
Last edited:

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
731
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

Well-known member
Jan 31, 2012
2,886
i don't see that we are getting consistence values for $c_1$ and $c_2$
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
731
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

Well-known member
Jan 31, 2012
2,886
but is that true for both y(0)=1 and y'(0)=0
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
731
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

Active member
Jan 8, 2013
255
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

Well-known member
Jan 31, 2012
2,886
ok that helps a lot.

we have probably massaged this problem enough🤔