Question

Find the general solution of the following differential equation (1 + Y square )+( x - e power tan inverse y)dy/dx=0

Answer 1

ANSWER

$Given \: question \: is \\ \\ (1 + y {}^{2} ) + (x - e {}^{ \tan {}^{ - 1} (y) } ) \frac{dy}{dx} = 0 \\ \\ \frac{(1 + y {}^{2}) }{(e {}^{ \tan {}^{ - 1} (y) } - x) } = \frac{dy}{dx} ...(01) \\ \\ put \: \: \: \tan {}^{ - 1} (y) = z \\ \\ Differentiate \: both \: sides \: w.r.t.y \\ \\ \frac{1}{1 + y {}^{2} } = \frac{dz}{dy} \\ \\ from \: equation \: 01 \\ \\ \frac{(1 + y {}^{2} )}{(e {}^{z} - x) } = \frac{dz(1 + y {}^{2} )}{dx} \\ \\ \frac{1}{e {}^{z} - x} = \frac{dz}{dx} \\ \\ \frac{dx}{dz} = e {}^{z} - 1 \\ \\ \frac{dx}{dz} + 1 = e {}^{z} \\ \\ \frac{(dx + dz)}{dz} = e {}^{z} \\ \\ d(x + z) = e {}^{z} dz \\ \\ Now \: take \: integral \: on \: both \: sides \: \\ \\ x + z = e {}^{z} + k \: \: \: \: ( \: k \: is \: constant) \\ \\ Now \: \: \: use \: of \: z \: (z = \tan {}^{ - 1} ( y ) ) \: in \:above \: equation \: \\ \\ x + \tan {}^{ - 1} (y) = e {}^{ \tan {}^{ - 1} (y) } + k \\ \\ Therefore \: the \: general \: solution \: of \: given \: differential \: equation \: is \: \\ x + \tan {}^{ - 1} (y) = e {}^{ \tan {}^{ - 1} (y) } + k \: \\ (where \: k \: is \: constant)$