Question

ODEs

I need to check some of my answers...

Answer 1

Is the solution of y'=-1.5y --> y=e^(1.5x)*c

Answer 2

Almost right, there's 2 mistakes

Answer 3

I will quickly type my steps...

Answer 4

\[y'=-1.5y\\dy=-1.5ydx\\\frac{1}{y}dy=-1.5dx\\ln(y)=-1.5x+C\\y=e^{-1.5x+C}\\y=e^{-1.5x}c\]

Answer 5

Now you're correct :), you forgot the minus just now, and why don't you write it
\[\huge y=e^{-1.5x}•e^c\]

Answer 6

Oh I see I forgot. Our lecturer told us to actually write it like that, since e^c is still a constant. But I'm glad I'm doing these problems correctly now

Answer 7

Yeah its fine but I won't write it like e^c=c, because It'll makes some confusion/mistakes if "c" was a variable, but that's rare

Answer 8

Oh yes because that does not equal c! Okay, I'll write it C_1, C_2, etc.

Answer 9

\[y'= \cosh (5.13x)\\ \rightarrow y=\int\limits{}{}\cosh(5.13x)dx\\y=\sinh(5.13x)\]?

Answer 10

forgot the +C

Answer 11

Did you integrate it correctly? :O

Answer 12

Try using u-sub
\[y=\int\limits \cosh(5.13x)dx \\ \\ \text{\let} \\ \\ u=5.13x \\ \\ du=5.13dx \\ \\ \frac{1}{5.13}du=dx \\ \\ y=\frac{1}{5.13}\int\limits \cosh(u)du \\ \\ y=\frac{1}{5.13}\sinh(u)+c \\ \\ y=\frac{1}{5.13}\sinh(5.13x)+c \]

Answer 13

Thanks, I'll remember that!

Answer 14

Now, this one was a little difficult for me..\[y'+4y=1.4\]


Is the solution\[y=1.4x-4yx+C\]?

Answer 15

I got \[\huge y=\frac{1.4}{4}-c_1e^{-4x}\]

Answer 16

I didn't really know what to do with this one, so I did this:
\[\frac{dy}{dx}+4y=1.4\\dy+4y dx=1.4 dx\\y=(1.4-4y)dx\\y=1.4x-4yx+C\]