Question

Solve the following differential equations given the initial value
dy/dx = tan [x], y[0] = 1

Answer 1

Answer:

                 y(x) = - log(cosx) + 1

Step-by-step explanation:

Given:

             $\frac{dy}{dx} = tan (x)$

             y = $\int\ {tanx} \, dx$

             $y=\int\ {\frac{sinx}{cosx} } \, dx$ ------------> Equation1         (Because $tanx=\frac{sinx}{cosx}$)

             Put   u = cosx

                     differentiate both side wrt x

                      $\frac{du}{dx}=-sinx$

                     - du = sinx dx

    Substitute -du in place of sinx dx and u = cosx in Equation 1

              $y=\int\ {\frac{-du}{u} } \, dx$

              y = - log(u) + C  (Because integration of 1/u is log(u))

               Now substitute u = cosx

            y(x) = - log(cosx) + C ---------------> Equation 2

            Given y(0) = 1

      Substitute x = 0 in Equation 2

         y(0) = -log(cos0) + C

              1 = -log(1) + C        (Because cos0 = 1)

              1 = 0 + C                (Because log(1)=0)

               C = 1

   From equation 1

                y(x) = - log(cosx) + C

                y(x) = - log(cosx) + 1

                                                    HOPE U UNDERSTOOD

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Answer 2

Given,

$\[\frac{{dy}}{{dx}} = \tan \left[ x \right],y\left[ 0 \right] = 1\]$

Solution,

Know that,

$\[\frac{{dy}}{{dx}} = \tan x\]$

Integrate with respect to x.

$\[y = - \ln \left( {\cos x} \right) + c\]$-------(1)

Where c is the constant.

Put initial value to get the value of constant.

$\[\begin{array}{l}1 = - \ln \left( {\cos 0} \right) + c\\ \Rightarrow 1 = - \ln 1 + c\\ \Rightarrow 1 = - 0 + c\\ \Rightarrow c = 1\end{array}\]$

Put it in equation 1.

$\[y = - \ln \left( {\cos x} \right) + 1\]$

Hence the solution of differential equation is $\[y = - \ln \left( {\cos x} \right) + 1\]$