1. ## Solving differential equations...

Hello, everyone. I am new here. This is just high school Calculus. Don't laugh at me .

Can someone tell me how these problems work? I even looked at the answers (yes, I *am* allowed to do that) but I can't work them backwards. The first one is this:

dy/dx = y+2

And this is what I did:

dy = (y+2) dx
y = Ce^x + 2x

Now the answer is y = Ce^x - 2 .

And the second one: y' = y√x

And last: (1 + x^2)y' = 2xy

2. ## Re: Solving differential equations...

Hello, italipinogirl!

$\frac{dy}{dx} \:=\: y+2$

And this is what I did: . $dy \:=\: (y+2) dx \quad\Rightarrow\quad y\:=\: Ce^x + 2x$
. . How did you get 2x?

Now the answer is: . $y \:=\: Ce^x - 2$

We have: . $\frac{dy}{dx} \:=\:y+2$

Separate variables: . $\frac{dy}{y+2} \:=\:dx$

Integrate: . $\int\frac{dy}{y+2} \:=\:\int dx \quad\Rightarrow\quad \ln|y+2| \:=\:x+c$

. . . . . . . . $y+2 \:=\:e^{x+c} \:=\:e^x\cdot e^c \:=\:e^x\cdot C$

. . . . . . . . $y+2 \:=\:Ce^x \quad\Rightarrow\quad y \:=\:Ce^x - 2$

$\frac{dy}{dx} \:=\: y\sqrt{x}$

We have: . $\frac{dy}{dx} \:=\:yx^{\frac{1}{2}}$

Separate variables: . $\frac{dy}{y} \:=\:x^{\frac{1}{2}}dx$

Integrate: . $\ln |y| \;=\;\tfrac{2}{3}x^{\frac{3}{2}}+c$

. . . . . . . . $y \;=\;e^{\frac{2}{3}x^{\frac{3}{2}} + c} \;=\;e^{\frac{2}{3}x^{\frac{3}{2}}}\cdot e^c \;=\;e^{\frac{2}{3}x^{\frac{3}{2}}}\cdot C$

. . . . . . . . $y \;=\;Ce^{\frac{2}{3}x^{\frac{3}{2}}}$

$(1 + x^2)\frac{dy}{dx}\:=\: 2xy$

Separate variables: . $\frac{dy}{y} \;=\;\frac{2x}{1+x^2}\,dx$

Integrate: . $\ln(y) \;=\;\ln(1+x^2) + c \;=\;\ln(1+x^2) + \ln C \;=\;\ln[C(1+x^2)]$

Therefore: . $y \;=\;C(1+x^2)$

3. ## Re: Solving differential equations...

First :
dy/dx = y+2
dy/(y+2) = dx
integrate it.

Second :
dy/dx = y√x
dy/y = (√x)dx
integrate it.

Third :
(1 + x^2)(dy/dx) = 2xy
dy/y = (2x/(1+x^2))dx
integrate it.

4. ## Re: Solving differential equations...

Wow, thank you! That completely makes sense now.