Question

i just need someone to check my differential equation

\[D^2y + 2D y + 50y = 0; \; y(0) = 7; \; y'(0) = 7\]

i got \[y= y_c = e^{-x} (7\cos 7x + 2\sin 7x)\]

Answer 1

Correct.

Answer 2

\[\Delta=b^2-4ac=2^2-4\times1\times50=-196\]\[m=\frac{-b\pm\sqrt{\Delta}}{2a}\]\[=\frac{-2\pm\sqrt{-196}}{2}\]\[m=-1\pm7i\]
\[y_c=e^{-x}(B\cos 7x+C\sin7x)\]

Answer 3

so...im right right?

Answer 4

\[y(0)=7\]
\[y(0)=e^{-0}(B\cos (7\cdot0)+C\sin(7\cdot0))\]\[7=B\cos (7\cdot0)\]\[7=B\]

Answer 5

uh-huh...

Answer 6

yes u are :)

Answer 7

\[y=e^{-x}(7\cos 7x+C\sin7x)\]\[y'=e^{-x}(-49\sin 7x+7C\cos7x) -e^{-x}(7\cos 7x+C\sin7x)\]\[y'=-e^{-x}((C+49)\sin 7x+7(C+1)\cos7x)\]
\[\qquad y'(0)=7\]
\[y'(0)=-e^{-0}((C+49)\sin (7\cdot0)+7(C+1)\cos(7\cdot 0))\]\[7=-7(C+1)\]?

Answer 8

negative/ o.O

Answer 9

i might have made a mistake