Question

Solve dy/dx - y tan x = 3e ^-sin x

Answer 1

Answer:

Step-by-step explanation:

dy/dx= y tanx +e^x

dy/dx- y tanx =e^x

P=-tanx → z=e^ ∫-( tanx) dx = e^∫ -tanx dx =e^ lncosx = cosx

dy/dx- y tanx =e^x

z=ycosx → dz=(dy*cosx-ysinxdx)

dy/dx- y tanx =e^x

( cosx)dy/dx- y tanx *cosx=e^x*cosx

( cosx)dy/dx- y sinx=e^x*cosx

( cosx)dy- y sinx dx=e^x*cosx dx

dz=e^x*cosx dx

z=∫e^x*cosx dx

z= 1/2 e^x (sin(x) + cos(x)) + constant

∫e^x*cosx dx - Wolfram|Alpha

z= 1/2 e^x (sin(x) + cos(x)) + constant

ycosx=1/2 e^x (sin(x) + cos(x)) + constant

y =1/2 e^x (tan(x) + 1) + constant/(cosx)

y(x) = C* sec(x) + (e^x)/2 + (1/2) e^x tan(x)

Answer 2

The General equation of the question is y(x) = 1/2 $e^x(sinx+cosx)+c$

Given:

dy/dx - y tan x = 3e ^-sin x which is in differential form

To Find:

The general equation of the following differential equation

Solution:

To calculate the integral factor ,

We should get the P and Q values,

P = tanx and Q = 3$e^{-sinx}$

Integral Factor (IF),

= e^∫tanxdx

= e^logsecx

IF = secx

To calculate the general solution,

We have,

y(IF) = ∫Q(IF)dx +c

ysecx = ∫3 e^-sinx(secx)dx +c

ysecx = cosx e^-sinx($\frac{1}{cosx)}$ + c

Upon cancelling the cos in numerator and denominator,

We get,

= $\frac{1}{2}e^x(sinx+cosx) +c$

Hence we get the requires solution as  $\frac{1}{2}e^x(sinx+cosx) +c$

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