Question

Differential Equation problem.

Answer 1

\[(\sqrt{x}+x) dy/dx = (\sqrt{y}+y)\]

Answer 2

I have tried different methods of integration but I might have to use substitution?

Answer 3

Just trying to find an explicit solution y =

Answer 4

yes try \(u = 1+\sqrt{y}\)

Answer 5

Would have to take out a multiple of \[\sqrt{y}\]

Answer 6

Yep !

Answer 7

\[\large (\sqrt{x}+x) \dfrac{dy}{dx} = (\sqrt{y}+y)\]

\[\large \dfrac{dy}{\sqrt{y}+y} = \dfrac{dx}{\sqrt{x}+x}\]

\[\large \dfrac{dy}{\sqrt{y}(1+\sqrt{y})} = \dfrac{dx}{\sqrt{x}(1+\sqrt{x})}\]

Answer 8

\[u = \sqrt{x} + 1\]
\[du = 1/\sqrt{x} dx\]

This step means i should take Sqrootx out of my integral?

Answer 9

careful derivative of \(\sqrt{x}\) is \(\dfrac{1}{2\sqrt{x}}\)

Answer 10

\[\large \int \dfrac{dy}{\sqrt{y}(1+\sqrt{y})} = \int \dfrac{dx}{\sqrt{x}(1+\sqrt{x})}\]

Answer 11

Right forgot the 1/2 was there

Answer 12

\(u = 1+\sqrt{y} \implies du = \dfrac{1}{2\sqrt{y}} dy \implies 2du = \dfrac{1}{\sqrt{y}} dy\)

Answer 13

\[\large \int \dfrac{2du}{u} = \int \dfrac{dx}{\sqrt{x}(1+\sqrt{x})}\]

Answer 14

work out the right hand side same way ^

Answer 15

Thank you very much it seems i was getting stuck with the algebra

Answer 16

\[2 \ln \left| 1+ \sqrt{x} \right| + c = 2 \ln \left| 1+ \sqrt{y} \right|\]

Answer 17

Looks good !

Answer 18

\[\ln(\frac{ 1+\sqrt{y} }{ 1+\sqrt{x} })^2 = c\]

Answer 19

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

Answer 20

Not sure on how to proceed to get this in terms y =

Answer 21

take sqrt both sides

Answer 22

\[e^c + x = y\]

Answer 23

?

Answer 24

and keep in mind c is as arbitrary as e^c is...

Answer 25

\[c_1 + x = \] Arbitrary C in mind

Answer 26

lets start from here :
\[2 \ln \left| 1+ \sqrt{x} \right| + c = 2 \ln \left| 1+ \sqrt{y} \right|\]

Answer 27

Not sure why that is yet?

Answer 28

Alright, lets not do that. keep c as it is :

\[\large 2 \ln \left| 1+ \sqrt{x} \right| + c = 2 \ln \left| 1+ \sqrt{y} \right| \]

\[\large \ln \left| 1+ \sqrt{x} \right| + \dfrac{c}{2} = \ln \left| 1+ \sqrt{y} \right| \]

\[\large e^{\ln \left| 1+ \sqrt{x} \right| + \frac{c}{2}} = 1+ \sqrt{y} \]

Answer 29

see if that looks okay so far ^

Answer 30

looks good except for the left side its still as e^ln

Answer 31

its okay, subtract 1 both sides and square both sides - you will have your y

Answer 32

after that, you can simplify if you want...

Answer 33

\[1+\sqrt{x} + e^{c/2} - 1 = \sqrt{y} \]

Answer 34

\[x + (e^{c/2})^2 = y\]
Would this be my final answer?

Answer 35

\[\large e^{\ln \left| 1+ \sqrt{x} \right| + \frac{c}{2}} = 1+ \sqrt{y}\]
\[\large e^{\ln \left| 1+ \sqrt{x} \right| + \frac{c}{2}}-1 = \sqrt{y}\]
\[\large \left(e^{\ln \left| 1+ \sqrt{x} \right| + \frac{c}{2}}-1\right)^2 = y\]

Answer 36

simplify the stuff inside parenthesis

Answer 37

\[\large \left(e^{\ln \left| 1+ \sqrt{x} \right| }e^{ \frac{c}{2}}-1\right)^2 = y\]

Answer 38

\[\large \left((1+\sqrt{x})e^{ \frac{c}{2}}-1\right)^2 = y\]

Answer 39

\[\large \left(A(1+\sqrt{x})-1\right)^2 = y\]

Answer 40

Oh I see where I made my mistakes again, Thank you very much!

Answer 41

np, you're welcome :)