Question

Question: In Problems , solve the equation.

$\left(\sqrt{\frac{y}{x}}\mathbf{+}\mathbf{cos}\mathbf{x}\right)\mathbf{dx}\mathbf{+}\left(\sqrt{\frac{x}{y}}\mathbf{+}\mathbf{sin}\mathbf{y}\right)\mathbf{dy}\mathbf{=}\mathbf{0}$

Short answer

$\mathbf{2}\mathbf{y}\sqrt{\frac{x}{y}}\mathbf{-}\mathbf{cos}\mathbf{y}\mathbf{+}\mathbf{sin}\mathbf{x}\mathbf{=}\mathbf{C}$

Step-by-step solution

Definition and concepts to be used 

Definition of Initial Value Problem:By an initial value problem for an nth-orderdifferential 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:

• Exactness:If $\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{=}\frac{\partial \mathbf{N}}{\partial \mathbf{x}}$. Otherwise, the equation is not exact.
• $\int {\mathbf{x}}^a\mathbf{dx}\mathbf{=}\frac{{\mathbf{x}}^{a+1}}{a+1}\mathbf{+}\mathbf{C}$.
• $\int \mathbf{sin}\mathbf{x}\mathbf{dx}\mathbf{=}\mathbf{-}\mathbf{cos}\mathbf{x}\mathbf{+}\mathbf{C}$.
• $\int \mathbf{cos}\mathbf{x}\mathbf{dx}\mathbf{=}\mathbf{sin}\mathbf{x}\mathbf{+}\mathbf{C}$.

Given information and simplification

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

Let us check whether the given equation is exact or not.

Then, $\mathbf{M}\mathbf{=}\sqrt{\frac{y}{x}}\mathbf{+}\mathbf{cos}\mathbf{x}\mathbf{,}\mathbf{N}\mathbf{=}\sqrt{\frac{x}{y}}\mathbf{+}\mathbf{sin}\mathbf{y}$.

Differentiate the value of M and N.

$\begin{array}{l}\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{=}\frac{1}{\mathbf{2}\sqrt{\mathrm{xy}}}\\ \frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{=}\frac{1}{\mathbf{2}\sqrt{\mathrm{xy}}}\end{array}$

So, the given equation is exact.

Evaluation method

Now, let us assume $\mathbf{N}\mathbf{=}\frac{\partial \mathbf{F}}{\partial \mathbf{y}}\mathbf{=}\sqrt{\frac{x}{y}}\mathbf{+}\mathbf{sin}\mathbf{y}$.

Integrate on both sides.

$\begin{array}{c}\mathbf{F}\mathbf{=}\int \left(\sqrt{\frac{x}{y}}\mathbf{+}\mathbf{sin}\mathbf{y}\right)\mathbf{dy}\\ \mathbf{=}\int \sqrt{\frac{x}{y}}\mathbf{dy}\mathbf{+}\int \mathbf{sin}\mathbf{y}\mathbf{dy}\\ \mathbf{=}\mathbf{2}\mathbf{y}\sqrt{\frac{x}{y}}\mathbf{-}\mathbf{cos}\mathbf{y}\mathbf{+}\mathbf{g}\left(x\right)\end{array}$

Differentiate the F with respect to x.

$\begin{array}{c}\frac{\partial \mathbf{F}}{\partial \mathbf{x}}\mathbf{=}\sqrt{\frac{y}{x}}\mathbf{+}\mathbf{g}\mathbf{\text{'}}\left(x\right)\\ =M\end{array}$

Equalise the values of M.

$\begin{array}{c}\sqrt{\frac{y}{x}}\mathbf{+}\mathbf{g}\mathbf{\text{'}}\left(x\right)\mathbf{=}\sqrt{\frac{y}{x}}\mathbf{+}\mathbf{cos}\mathbf{x}\\ \mathbf{g}\mathbf{\text{'}}\left(x\right)\mathbf{=}\mathbf{cos}\mathbf{x}\end{array}$

Integrate on both sides.

$\begin{array}{c}\int \mathbf{g}\mathbf{\text{'}}\left(x\right)\mathbf{=}\int \mathbf{cos}\mathbf{x}\mathbf{dx}\\ \mathbf{g}\left(x\right)\mathbf{=}\mathbf{sin}\mathbf{x}\mathbf{+}{\mathbf{C}}_1\end{array}$

Substitute in equation of F.

$\begin{array}{c}\mathbf{2}\mathbf{y}\sqrt{\frac{x}{y}}\mathbf{-}\mathbf{cos}\mathbf{y}\mathbf{+}\mathbf{sin}\mathbf{x}\mathbf{+}{\mathbf{C}}_1\mathbf{=}\mathbf{0}\\ \mathbf{2}\mathbf{y}\sqrt{\frac{x}{y}}\mathbf{-}\mathbf{cos}\mathbf{y}\mathbf{+}\mathbf{sin}\mathbf{x}\mathbf{=}\mathbf{C}\end{array}$

Hence, the solution of the given initial value problem is

$\mathbf{2}\mathbf{y}\sqrt{\frac{x}{y}}\mathbf{-}\mathbf{cos}\mathbf{y}\mathbf{+}\mathbf{sin}\mathbf{x}\mathbf{=}\mathbf{C}$.