Differential Equations help: verify that...

Q: Verify that the given relation is an implicit solution for the given problem.

Problem:

y dy/dx = 2x(1+y) , y = x^2 + ln(1+y)

which is in the book.

So obviously, it's separable.

(ydy)/(1+y) = 2x dx

The right side is cake when taking the integral... its the left side thats messing me up and causing me to not get the right answer, when solving for y.

Left side is integration by parts... which is what I've tried. Any help on getting to that implicit answer would be appreciated. THanks.

You're off to a good start, but it's difficult to get y alone on one side.

\(\displaystyle \L\\y\frac{dy}{dx}=2x(1+y)\)

Separate, as you have done:

\(\displaystyle \L\\\frac{y}{1+y}dy=2xdx\)

Partial fractions on left side and integrate:

\(\displaystyle \L\\\int{1}dy-\int\frac{1}{y+1}dy=\int{2x}dx\)

\(\displaystyle \L\\y-ln(y+1)=x^{2}\)

\(\displaystyle \L\\y=x^{2}+ln(y+1)\)

Hello, sekira!

You're making too much work out of it . . .

Verify that \(\displaystyle \,y \:=\:x^2\,+\,\ln(1\,+\,y)\,\) is an implicit solution for: \(\displaystyle \,y\,\frac{dy}{dx}\;=\;2x(1\,+\,y)\)

Click to expand...

It does not ask you to solve the differential equation.

They already did that and gave you the answer
. . You are asked to verify that it is indeed the right answer.

[It is similar to: Verify that \(\displaystyle x\,=\,3\) is a solution of: \(\displaystyle \,5x\,-\,4\:=\:3x\,+\,2\)
. . You just "plug in", right?]


We have: \(\displaystyle \,y\:=\:x^2\,+\,\ln(1\,+\,y)\)

Then: \(\displaystyle \;\frac{dy}{dx}\:=\:2x\,+\,\frac{1}{1\,+\,y}\left(\frac{dy}{dx}\right)\;\;\Rightarrow\;\;\frac{dy}{dx}\,-\,\frac{1}{1\,+\,y}\left(\frac{dy}{dx}\right)\;=\;2x\)

Factor: \(\displaystyle \,\frac{dy}{dx}\left(1\,-\,\frac{1}{1\,+\,y}\right)\:=\:2x\;\;\Rightarrow\;\;\frac{dy}{dx}\left(\frac{y}{1\,+\,y}\right)\:=\:2x\;\;\Rightarrow\;\;\frac{dy}{dx}\:=\:\frac{2x(1\,+\,y)}{y}\)


Substitute into the differential equation:

. . . . \(\displaystyle y\,\frac{dy}{dx}\;\;\;=\;2x(1\,+\,y)\)
. . . . . . \(\displaystyle \downarrow\)
. . \(\displaystyle y\,\overbrace{\frac{2x(1\,+\,y)}{y}} \;= \;2x(1\,+\,y)\)

. . .\(\displaystyle 2x(1\,+\,y) \;=\;2x(1\,+\,y)\) . . . There!

Thanks both of you.

Yeah... either way would work. I was just used to solving them b/c of the previous 8 problems... so I didn't even bother to think of plugging in and just checking that way. But both ways worked... in about the same amount of time.