# Integrate : dy/dx = 2x + y ?

Apr 1, 2018

${x}^{2} + x y + c$

#### Explanation:

=$\int \mathrm{dy} = \int 2 x + y \mathrm{dx}$

=$y = \frac{2 {x}^{2}}{2} + x y + c$

=$y = {x}^{2} + x y + c$

Hope it helps!

Apr 1, 2018

$\text{ The GS is, } y \cdot {e}^{-} x + 2 \left(x + 1\right) {e}^{-} x = C , \mathmr{and} ,$

$y + 2 \left(x + 1\right) = C {e}^{x}$.

#### Explanation:

Rewriting the given diff. eqn. (DE) as $\frac{\mathrm{dy}}{\mathrm{dx}} - y = 2 x$, we find

that it is a linear DE of the form : $\frac{\mathrm{dy}}{\mathrm{dx}} + y P \left(x\right) = q \left(x\right)$.

To find its gen. soln. (GS), we need to multiply it by the

integrating factor (IF) e^(intP(x)dx.

Since,

$P \left(x\right) = - 1 , \int P \left(x\right) \mathrm{dx} = \int - 1 \mathrm{dx} = - x \therefore \text{ IF is } {e}^{-} x .$

Multiplying the DE by IF, we get,

${e}^{-} x \frac{\mathrm{dy}}{\mathrm{dx}} - y {e}^{-} x = 2 x {e}^{-} x$.

$\therefore {e}^{-} x \cdot \frac{d}{\mathrm{dx}} \left(y\right) + y \cdot \frac{d}{\mathrm{dx}} \left({e}^{-} x\right) = 2 x {e}^{-} x , \mathmr{and} ,$

$\frac{d}{\mathrm{dx}} \left(y \cdot {e}^{-} x\right) = 2 x {e}^{-} x$.

$\therefore y \cdot {e}^{-} x = \int 2 x {e}^{-} x \mathrm{dx} + C$,

=2[x*inte^-xdx-int{d/dx(x)inte^-xdx}dx]+C......[because," Integration by Parts]",

$= 2 \left[x \left(- {e}^{-} x\right) - \int \left(- {e}^{-} x\right) \mathrm{dx}\right] + C$,

$= 2 \left[- x {e}^{-} x - {e}^{-} x\right] + C$.

$\Rightarrow \text{ The GS is, } y \cdot {e}^{-} x + 2 \left(x + 1\right) {e}^{-} x = C , \mathmr{and} ,$

$y + 2 \left(x + 1\right) = C {e}^{x}$.

Feel the Joy of Maths.!