Question

\[y''+3y=0\]
\[y(0)=1\]
\[y'(0)=3\]
\[y(0)=c_1cos\sqrt3(0)-c_2sin\sqrt3(0)=1\]

Answer 1

Do you know what \(\cos(0)\) and \(\sin(0)\) are?

Answer 2

yes, cos(0)=1

Answer 3

why not just find the derivative of y(x)?

Answer 4

\[c_2\sin\sqrt({3}*0)=0\]
leaves
\[c_1\cos(\sqrt{3}*0)=1\]
\[c_1*1=1\]
\[c_1=1\]

\[y(x)=c_1\cos(x\sqrt{3})-c_2\sin(x\sqrt{3})\]
\[y'(x)=-c_1\sqrt{3}sin(x\sqrt{3})-c_2\sqrt{3}cos(x\sqrt{3})\]

i hope i didnt derive it incorrectly
anyways
\[f'(0)=3=-c_1\sqrt{3}sin(0\sqrt{3})-c_2\sqrt{3}cos(0\sqrt{3})\]
substitute in c_1 and solve for c_2

Answer 5

Thank you!

Answer 6

so would \[c_2=\frac{-3}{\sqrt3}\]