(1+tanx)dy=2siny dx solve differential equation?

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Answer 1 · 2017-09-18T10:08:42

$y = 2 a r c tan {c (sin x + cos x) e^{x}},$ $y=2arc tan{c(sinx+cosx)e^x},$ is the GS.

Rewriting the given Diff. Eqn., as,

$\frac{d y}{sin y} = \frac{2}{1 + tan x} d x,$ $dy/siny=2/(1+tanx)dx,$ we find that, it is Separable Variable type.

$:. cscydy=(2cosx)/(sinx+cosx)dx.$

The General Solution (GS) is obtained by integrating term-wise.

$:. intcsc y dy=int(2cosx)/(sinx+cosx)dx+lnc.$

$:. ln(tan(y/2))=int{(sinx+cosx)+(cosx-sinx)}/(sinx+cosx)dx+lnc.$

$=int{(sinx+cosx)/(sinx+cosx)+(cosx-sinx)/(sinx+cosx)}dx+lnc,$

$=int1dx+int1/tdt,$

$where, sinx+cosx=t rArr (cosx-sinx)dx=dt.$

$:. ln(tan(y/2))=x+lnt+lnc, i.e., x+ln(sinx+cosx)+lnc.$

$:. ln(tan(y/2))-ln(sinx+cosx)=x+lnt, or,$

$ln{(tan(y/2))/(sinx+cosx)}=lne^x+lnc=ln(ce^x),$

$:. tan(y/2)/(sinx+cosx)=ce^x.$

$:. tan(y/2)=c(sinx+cosx)e^x.$

$:. y/2=arc tan{c(sinx+cosx)e^x}.$

$rArr y=2arc tan{c(sinx+cosx)e^x},$ is the desired GS.

Enjoy Maths.!