# [SOLVED]1.8.17 initial value then find C

#### karush

##### Well-known member
$\tiny{1.8.17}$
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
What are the directions asking you to do? What are the second derivatives of $\cos(2t)$ and $\sin(2t)$?

#### karush

##### Well-known member
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
The problem said "use direct substitution"! So substitute those into the given equation.

#### karush

##### Well-known member
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
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
i don't see that we are getting consistence values for $c_1$ and $c_2$

#### Country Boy

##### Well-known member
MHB Math Helper
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
but is that true for both y(0)=1 and y'(0)=0

#### Country Boy

##### Well-known member
MHB Math Helper
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
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
ok that helps a lot.

we have probably massaged this problem enough