Question

d^2y/dx^2=9y solve for y

Answer 1

let \(\huge \frac{dy}{dx}=p \) and note that \(\huge \frac{dp}{dx}=\frac{dp}{dy}\frac{dy}{dx}=\frac{dp}{dy} p \)

Answer 2

i know that the answer is e^(-3x) or e^(3x) im just not sure about the steps

Answer 3

now simply separate the variables
\[\frac{\text d^2y}{\text dx^2}=9y\]
\[\frac{\text dp}{\text dy}p=9y\]
\[p{\text dp}=9y\text dy\]
and integrate
\[\int p\cdot{\text dp}=9\int y\cdot\text dy\]

Answer 4

\[\frac{p^2}2=9\frac{y^2}2+c_1\]\[{p^2}=9y^2+c_2\]
\[p=\sqrt{9y^2+c_2}\]
\[\frac{\text dy}{\text dx}=\sqrt{9y^2+c_2}\]
\[\frac{\text dy}{\sqrt{9y^2+c_2}}=\text dx\]
\[\frac 13\int\frac{\text dy}{\sqrt{y^2+c_2}}=\int\text dx\]

Answer 5

that should be \(c_3\) in the last equation, \(c_2/9=c_3\)

Answer 6

thanks!

Answer 7

There is an easier method;\[\frac{\text d^2y}{\text dx^2}=9y\]\[\frac{\text d^2y}{\text dx^2}-9y=0\]\[m^2-9=0\]\[(m+3)(m-3)=0\]\[m=\pm 3\]

Answer 8

Thanks that is easier to follow

Answer 9

If it was -9y rather than 9y would your only option be to use the first method

Answer 10

\[\frac{\text d^2y}{\text dx^2}=-9y\]\[\frac{\text d^2y}{\text dx^2}+9y=0\]\[m^2+9=0\]\[(m+3i)(m-3i)=0\]\[m=\pm 3i\]

Answer 11

which correlates to sin3x or cos3x?

Answer 12

\[\frac{\text d^2y}{\text dx^2}=9y\qquad\Longrightarrow\qquad y(x)=Ae^{3x}+B^{-3x}\]

\[\frac{\text d^2y}{\text dx^2}=-9y\qquad\Longrightarrow\qquad y(x)=A\cos{3x}+\sin{3x}\]

Answer 13

ops there should be a constant on the last term of the last line \(\cdots+B\sin3x\)

Answer 14

great! thanks again!