Question

differential equations, Y(partial)(t)=ln(2+4t), t>-1/2, is a particular solution to the differential equation y''+8y=g(t). find g(t). HELP

Answer 1

What does Y partial t look like in regular mathical notations ...

Answer 2

\[Y _{p}(t)\]

Answer 3

hmm, seems like my latex coder is still on the fritz.

but it seems like to me that if we let y = ln(2+4t), then all we have to do is determine y'' and fill it all in

Answer 4

y = y_p(t) + y_h(t), if memory serves

Answer 5

y'' + 8y = 0 gives us y_h

can you recall how to find this if need be with a characteristic setup?

Answer 6

\[y(x)=(c1)\cos(2\sqrt{2}x)+(c2)\sin(2\sqrt{2}x)+\ln(2+4t)\]

Answer 7

i agree with that :)

Answer 8

t instead of x tho

Answer 9

i know the solution but i need to find the original differential equation from which it came in the for y''+8y=g(t) I don't know how to find g(t).

Answer 10

if you know y, then determine y'' and 8y, and add them together

Answer 11

g(t) is equal to (y'' + 8y), and y=y_p + y_h

Answer 12

that does make the most sense right?

Answer 13

y=(c1) cos(2sqrt{2} t) + (c2) sin(2sqrt{2} t) + ln(2+4t)

y'=-2sqrt(2)(c1) sin(2sqrt{2} t) + 2sqrt(2)(c2) cos(2sqrt{2} t) + 4(2+4t)^(-1)

y''=-8(c1) cos(2sqrt{2} t) - 8(c2) sin(2sqrt{2} t) - 16(2+4t)^(-2)

Answer 14

I think "p" means "particular" in this case, not partial.

Answer 15

i learned it with y_c and y_h :)

Answer 16

8y = 8(c1) cos(2sqrt{2} t) + 8(c2) sin(2sqrt{2} t) + 8ln(2+4t)
+y''=-8(c1) cos(2sqrt{2} t) - 8(c2) sin(2sqrt{2} t) - 16(2+4t)^(-2)
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g(t) = 0 + 0 +8ln(2+4t) -16/(2+4t)^2
|the homogenous part as expected|