# Thread: Ordinary differentiation equations (ODE)-seperable

1. ## Ordinary differentiation equations (ODE)-seperable

hey there..
i've got unsolved solution here..

Solve initial value problem..

xy  y = 2x2y ; y(1) = 1

x dy/dx = y + 2x2y
dy/dx = y/x + 2xy
dy/dx = y[(1/x) + 2x]
∫1/y dy = ∫ (1/x) + 2x dx
ln |y| = ln |x| + x2 + C
apply exp to both side
y = x + ex^2 + e^C

substitute the initial value
1 = 1 + e1 + e^C
e^(1+C) = 0
apply ln
1 + C = ln 0?

2. Hello, nameck!

We're supposed to guess which numbers are exponents?

$\displaystyle xy - y \:=\: 2x^2y\qquad y(1) \,=\, 1$

We have: .$\displaystyle x\,\frac{dy}{dx} \;=\;2x^2y + y$

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

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

. . . . . . . . .$\displaystyle \frac{dy}{y} \;=\;\left(2x + \frac{1}{x}\right)\,dx$

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

. . . . . . . . . .$\displaystyle y \;=\;e^{x^2+\ln x+C}$

. . . . . . . . . .$\displaystyle y \;=\;e^{x^2}\cdot e^{\ln x}\cdot e^C$

. . . . . . . . . .$\displaystyle y \;=\;e^{x^2}\cdot x \cdot C$

. . Hence: . . $\displaystyle y \;=\;Cx\,e^{x^2}$

Since $\displaystyle y(1)=1$, we have: .$\displaystyle C\cdot 1\cdot e^{1^2} \:=\:1 \quad\Rightarrow\quad C \:=\:\frac{1}{e}$

Therefore: .$\displaystyle y \;=\;\frac{1}{e}\,x\,e^{x^2} \quad\Rightarrow\quad\boxed{ y\;=\;x\,e^{x^2-1}}$