Integrate : dy/dx = 2x + y ?

Question

15378 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-01T18:38:22

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

= $\int d y = \int 2 x + y d x$ $intdy=int2x+ydx$

= $y = \frac{2 x^{2}}{2} + x y + c$ $y=(2x^2)/2+xy+c$

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

Hope it helps!

Answer 2 · 2018-04-01T18:58:42

$The GS is, y \cdot e^{- x} + 2 (x + 1) e^{- x} = C, or,$ $" The GS is, "y*e^-x+2(x+1)e^-x=C, or,$

$y + 2 (x + 1) = C e^{x}$ $y+2(x+1)=Ce^x$ .

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

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

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

integrating factor (IF) $e^{\int P (x) d x}$ $e^(intP(x)dx$ .

Since,

$P(x)=-1, intP(x)dx=int-1dx=-x :." IF is "e^-x.$

Multiplying the DE by IF, we get,

$e^-xdy/dx-ye^-x=2xe^-x$ .

$:. e^-x*d/dx(y)+y*d/dx(e^-x)=2xe^-x, or,$

$d/dx(y*e^-x)=2xe^-x$ .

$:. y*e^-x=int2xe^-xdx+C$ ,

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

$=2[x(-e^-x)-int(-e^-x)dx]+C$ ,

$=2[-xe^-x-e^-x]+C$ .

$rArr" The GS is, "y*e^-x+2(x+1)e^-x=C, or,$

$y+2(x+1)=Ce^x$ .

Feel the Joy of Maths.!