Solve the differential equation $(dy)/(dx) + x tan(y+x) = 1$ ?

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Answer 1 · 2016-06-04T10:16:18

One solution is $y(x) = x + arcsin(e^{-1/2x^2+C})$

First we make the variable change $z(x) = y(x)-x$ and

$(dz)/(dx) = (dy)/(dx)-1$

substituting

$1 + (dz)/(dx) + x tan(z) = 1 equiv (dz)/(dx) + x tan(z) = 0$

grouping

$(dz)/(tan(z))=-x dx equiv d/(dx)(log_e(sin(z)))=-d/dx(1/2x^2)$
$log_e sin(z) = -1/2 x^2+ C->sin(z) = e^{-1/2x^2+C}$

and finally

$z(x) = y(x)-x = arcsin(e^{-1/2x^2+C})$

then

$y(x) = x + arcsin(e^{-1/2x^2+C})$

Solve the differential equation (dy)/(dx) + x tan(y+x) = 1 (dy)/(dx) + x tan(y+x) = 1 ?