Question

Differential Equation problem.

Answer 1

\[dy/dx = e^\sqrt{x}/y\]

Answer 2

Completely lost on this next one i have separated the variables and do not know what to do with \[e^\sqrt{x} dx\]

Answer 3

It also is an IVP y(1) = 4

Answer 4

Do a substitution: put \[\sqrt{x} = t\] and so dx = 2tdt now do the required steps

Answer 5

\[\sqrt{x} = t\]
\[x = t^2\]
\[dx = 2t dt\]
\[e^t dx = y \times dy\]
\[\int\limits_{}^{} e^t 2tdt = ?\]

Answer 6

That's right. Now you can use integration by parts. Do you need any help for integration by parts ?

Answer 7

\[UV-\int\limits_{}^{}v dU\]
\[U = e^t\]

Answer 8

\[dU= e^t dt\]
\[dV = 2t dt\]
\[V = t^2\]
\[e^t \times t^2 - \int\limits_{}^{} t^2 e^t dt\]

Answer 9

Not sure from here

Answer 10

try

u = 2t dv = e^t

Answer 11

Yes. Do as dumbcow says

Answer 12

\[U = 2t \]
\[dU = 2 \times dt\]
\[dV = e^t dt\]
\[V = e^t\]
\[2t \times e^t - \int\limits_{}^{}e^t \times 2 dt\]

Answer 13

\[2t \times e^t -2e^t\]

Answer 14

That's right.

Answer 15

\[2\sqrt{x} \times e^\sqrt{x}-2e^\sqrt{x}\]?

Answer 16

yes.\[2(\sqrt{x}-1)e ^{\sqrt{x}}\]

Answer 17

\[\frac{ y^2 }{ 2 } = 2\sqrt{x} \times e^\sqrt{x}- 2 e^\sqrt{x} +c\]

Answer 18

\[y = \sqrt{4(\sqrt{x}-1)e^\sqrt{x}+c}\]

Answer 19

correct, now plug in initial point (1,4) to find c

Answer 20

\[4 = 2 \sqrt{e^\sqrt{1}(\sqrt{1}-1)+c}\]

Answer 21

\[4 = 2\sqrt{e^\sqrt{1}(1-1)+c}\]
\[4 = 2\sqrt{c}\]
\[2 = \sqrt{c}\]
\[4 = c\]

Answer 22

\[y = 2\sqrt{e^\sqrt{x}(\sqrt{x}-1)+4}\]

Answer 23

correct :)

Answer 24

NICE, Thank you very much