We #160;have,dydx=x+y2Let #160;x+y=v #8658;1+dydx=dvdx #8658;dydx=dvdx 1 #8756;dvdx 1=v2 #8658;dvdx=v2+1 #8658;1v2+1dv=dxIntegrating #160;both ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Integration to Solve Modified Sum of Binomial Coefficients

d y d x=x+y 2

Question

$\frac{d y}{d x} = {(x + y)}^{2}$

Open in App

Solution

$We have, \frac{d y}{d x} = {(x + y)}^{2} Let x + y = v \Rightarrow 1 + \frac{d y}{d x} = \frac{d v}{d x} \Rightarrow \frac{d y}{d x} = \frac{d v}{d x} - 1 ∴ \frac{d v}{d x} - 1 = v^{2} \Rightarrow \frac{d v}{d x} = v^{2} + 1 \Rightarrow \frac{1}{v^{2} + 1} d v = d x Integrating both sides, we get \int \frac{1}{v^{2} + 1} d v = \int d x \Rightarrow \tan^{- 1} v = x + C \Rightarrow v = \tan (x + C) \Rightarrow x + y = \tan (x + C)$

Suggest Corrections

Similar questions

sin (x y) + \frac{y}{x} = x^{2} - y^{2}

\frac{d y}{d x}

\frac{d y}{d x} = \frac{x + x^{2}}{y - y^{2}}

y = e^{x} + e^{- x}

\frac{d y}{d x} = \sqrt{y^{2} - 4}

\frac{d y}{d x} = y \tan x, y (0) = 1

2 x \frac{d y}{d x} = 5 y, y (1) = 1

\frac{d y}{d x} = 2 e^{2 x} y^{2}, y (0) = - 1

\cos y \frac{d y}{d x} = e^{x}, y (0) = \frac{π}{2}

\frac{d y}{d x} = 2 x y, y (0) = 1

\frac{d y}{d x} = 1 + x^{2} + y^{2} + x^{2} y^{2}, y (0) = 1

x y \frac{d y}{d x} = (x + 2) (y + 2), y (1) = - 1

\frac{d y}{d x} = 1 + x + y^{2} + x y^{2}

2 (y + 3) - x y \frac{d y}{d x} = 0

\frac{d y}{d x} = y \tan x, y (0) = 1

2 x \frac{d y}{d x} = 5 y, y (1) = 1

\frac{d y}{d x} = 2 e^{2 x} y^{2}, y (0) = - 1

\cos y \frac{d y}{d x} = e^{x}, y (0) = \frac{π}{2}

\frac{d y}{d x} = 2 x y, y (0) = 1

\frac{d y}{d x} = 1 + x^{2} + y^{2} + x^{2} y^{2}, y (0) = 1

x y \frac{d y}{d x} = (x + 2) (y + 2), y (1) = - 1

Join BYJU'S Learning Program