Question

dy/dx+ycotx=e^cosx solve diffrrential eqn​

Answer 1

Given:

•    $\frac{dy}{dx} + y \ cotx = e^{cosx}$

To Find:

•    Solution of the given differential equation.

Solution:

     $\frac{dy}{dx} + y \ cotx = e^{cosx}$

The given differential equation in the form of

     ⇒    $\frac{dy}{dx} + P(x)y = Q(x)$

     ⇒    $I F = e^ \int^ P$

     ⇒    $IF = e^\int^{cot x}$

     ⇒    $IF = e^{log \ sinx}$

     ⇒   $IF = sin x$

Now consider,

   ⇒    $y \times IF = \int IF \times Q(x) dx$

   ⇒    $y \times sinx = \int IF \times e^{cosx} dx$

  Let' s assume

         ⇒     $cos x = t$

         ⇒     $-sinx \ dx = dt$

Now,

  ⇒     $y \times sinx = - \int e^tdt$

  ⇒    $y \times sinx = - e^t + c$

Substituting the value of t

          $y \ sinx = - e^{cosx} + c$

Thus,

        $y \ sinx = - e^{cosx} + c$