Question

solve diffrrential eqn dy /dx +ycotx =e^cosx at x=pi/2,y=-2​

Answer 1

The solution of diffrrential eqn dy /dx +ycotx =e^cosx at x=pi/2,y=-2​ is as follows:

Given,

dy /dx + ycotx = e^cosx

dy /dx + P(x)y = Q(x)

$IF = e^{\int P dx}\\\\= e^{\int cotx dx}\\\\= e^{log sinx}\\\\= sin x$

Therefore, IF = sinx

Now consider,

y × IF = ∫ IF × Q(x) dx

$y \times sinx = \int e^{cosx}sinx dx\\\\let, cosx =t , -sinx dx = dt\\\\ysinx = -\int e^t dt\\\\ysinx = -e^t + c\\\\ysinx = -e^{cosx}+c$

Given, x = pi/2, y = -2​

y sin x = -e^{cos x} + c

(-2) × sin (π/2) = - e^{cos (π/2)} + c

-2 × 1 = -e^{0} +c

-2 = -1 + c

-2 + 1 = c

c = -1

y sin x  = -e^{cos x} - 1

is the required solution.