How to solve for the particular solution of the separable Differential equation

[tops]

1. 2xydx+(1+y)dy=0 ; y(2)=1

2. y'+y^2 sinx=0 ; y(0)=0

3. 2xdx-dy=x(xdy-2ydx) ; y(-3)=1

4. dy=x(2ydx-xdy) ; x=1 , y=4

 

[tkhunny]

You should start by separating, since you know they are separable.

Let's see what you get.

 

[soroban]

Hello, tops!

You've never done one of these? . . . Ever?
I'll do the first one.

\(\displaystyle (1)\;\;2xy\,dy + (1+y)dy \;=\; 0 \quad y(2) = 1 \)

Separate variables: .[\(\displaystyle \frac{y+1}{y}\,dy \;=\;2x \,dx\)

Integrate: .\(\displaystyle \displaystyle\int\left(1 + \tfrac{1}{y}\right)\,dy \;=\;\int 2x\,dx\)

. . . . . . . . . . . . .\(\displaystyle y + \ln y \;=\;x^2 + C\)


\(\displaystyle y(2) = 1\!:\;\; 1 + \ln 1 \:=\:2^2+C \quad\Rightarrow\quad C \,=\,-3\)


\(\displaystyle \text{Therefore: }\;y + \ln y \;=\;x^2 - 3\)