Question: In Problems 1-30, solve the equation.

$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{+}\mathbf{y}\mathbf{tan}\mathbf{x}\mathbf{+}\mathbf{sin}\mathbf{x}\mathbf{=}\mathbf{0}$

Definition of Initial Value Problem: By an initial value problem for an nth-order differential equation $\mathbf{F}\left(\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{,}\mathrm{...}\mathbf{,}\frac{{\mathbf{d}}^n\mathbf{y}}{{\mathbf{dx}}^n}\right)\mathbf{=}\mathbf{0}$ we mean: Find a solution to the differential equation on an interval I that satisfies at x₀ the n initial conditions

$\begin{array}{c}\mathbf{y}\left({\mathbf{x}}_0\right)\mathbf{=}{\mathbf{y}}_0\mathbf{,}\\ \frac{\mathrm{dy}}{\mathrm{dx}}\left({\mathbf{x}}_0\right)\mathbf{=}{\mathbf{y}}_1\mathbf{,}\\ .\\ .\\ .\\ \frac{{\mathbf{d}}^{n-1}\mathbf{y}}{{\mathbf{dx}}^{n-1}}\left({\mathbf{x}}_0\right)\mathbf{=}{\mathbf{y}}_{n-1}\mathbf{,}\mathbf{,}\end{array}$

Where${\mathbf{x}}_0\in \mathbf{I}$ and${\mathbf{y}}_0\mathbf{,}{\mathbf{y}}_1\mathbf{,}\mathrm{...}\mathbf{,}{\mathbf{y}}_{n-1}$ are given constants.

Formulae to be used:

• Integration by parts:.$\int \mathbf{udv}\mathbf{=}\mathbf{uv}\mathbf{-}\int \mathbf{vdu}$
• $\int {\mathbf{x}}^a\mathbf{dx}\mathbf{=}\frac{{\mathbf{x}}^{a+1}}{a+1}\mathbf{+}\mathbf{C}$.
• $\int \frac{1}{x}\mathbf{dx}\mathbf{=}\mathbf{ln}\left|x\right|\mathbf{+}\mathbf{C}$
  
• $\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{+}\mathbf{P}\left(x\right)\mathbf{y}\mathbf{=}\mathbf{Q}\left(x\right)$
  

Given that,$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{+}\mathbf{y}\mathbf{tan}\mathbf{x}\mathbf{+}\mathbf{sin}\mathbf{x}\mathbf{=}\mathbf{0}\mathrm{......}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Let $\mathbf{P}\left(x\right)\mathbf{=}\mathbf{tan}\mathbf{x}$.

Find the value of $\mu \left(x\right)$.

$\begin{array}{c}\mu \left(x\right)\mathbf{=}{\mathbf{e}}^{\int \mathbf{P}\left(x\right)\mathbf{dx}}\\ \mathbf{=}{\mathbf{e}}^{\int \mathbf{tan}\mathbf{xdx}}\\ \mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{ln}\left|\mathbf{cos}\mathbf{x}\right|}\\ \mathbf{=}\frac{1}{\mathbf{cos}\mathbf{x}}\end{array}$

Multiply $\frac{1}{\mathbf{cos}\mathbf{x}}$in equation (1) on both sides.

$\begin{array}{c}\frac{1}{\mathbf{cos}\mathbf{x}}\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{+}\mathbf{y}\mathbf{tan}\mathbf{x}\frac{1}{\mathbf{cos}\mathbf{x}}\mathbf{=}\mathbf{-}\frac{\mathbf{sin}\mathbf{x}}{\mathbf{cos}\mathbf{x}}\\ \frac{d}{\mathrm{dx}}\left[\frac{1}{\mathbf{cos}\mathbf{x}}\mathbf{y}\right]\mathbf{=}\mathbf{-}\frac{\mathbf{sin}\mathbf{x}}{\mathbf{cos}\mathbf{x}}\end{array}$

Now integrate the equation on both sides.

$\begin{array}{c}\int \frac{d}{\mathrm{dx}}\left[\frac{1}{\mathbf{cos}\mathbf{x}}\mathbf{y}\right]\mathbf{dx}\mathbf{=}\mathbf{-}\int \frac{\mathbf{sin}\mathbf{x}}{\mathbf{cos}\mathbf{x}}\mathbf{dx}\\ \frac{1}{\mathbf{cos}\mathbf{x}}\mathbf{y}\mathbf{=}\mathbf{-}\int \frac{\mathbf{sin}\mathbf{x}}{\mathbf{cos}\mathbf{x}}\mathbf{dx}\mathrm{.....}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\end{array}$

Find the value of$\mathbf{-}\int \frac{\mathbf{sin}\mathbf{x}}{\mathbf{cos}\mathbf{x}}\mathbf{dx}$separately.

Let us take $\mathbf{u}\mathbf{=}\mathbf{cos}\mathbf{x}$and $\mathbf{dx}\mathbf{=}\mathbf{-}\frac{1}{\mathbf{sin}\mathbf{x}}\mathbf{du}$.

$\begin{array}{c}\mathbf{-}\int \frac{\mathbf{sin}\mathbf{x}}{\mathbf{cos}\mathbf{x}}\mathbf{dx}\mathbf{=}\int \frac{1}{u}\mathbf{du}\\ \mathbf{=}\mathbf{ln}\left|u\right|\mathbf{+}\mathbf{C}\\ \mathbf{=}\mathbf{ln}\left|\mathbf{cos}\mathbf{x}\right|\mathbf{+}\mathbf{C}\end{array}$

Now substitute in equation (2)

$\begin{array}{c}\frac{1}{\mathbf{cos}\mathbf{x}}\mathbf{y}\mathbf{=}\mathbf{ln}\left|\mathbf{cos}\mathbf{x}\right|\mathbf{+}\mathbf{C}\\ \mathbf{y}\mathbf{=}\mathbf{cos}\mathbf{x}\mathbf{ln}\left|\mathbf{cos}\mathbf{x}\right|\mathbf{+}\mathbf{C}\mathbf{cos}\mathbf{x}\end{array}$

Hence, the solution of the given initial value problem is

$\mathbf{y}\mathbf{=}\mathbf{cos}\mathbf{x}\mathbf{ln}\left|\cos x\right|\mathbf{+}\mathbf{C}\mathbf{cos}\mathbf{x}$