91Ó°ÊÓ

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${\mathbf{x}}_0∈\mathbf{I}$ 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: $∫\mathbf{udv}\mathbf{=}\mathbf{uv}\mathbf{-}∫\mathbf{vdu}$.
• $∫{\mathbf{e}}^{\mathrm{ax}}\mathbf{dx}\mathbf{=}\frac{{\mathbf{e}}^{\mathrm{ax}}}{a}\mathbf{+}\mathbf{C}$.
• Product rule: $\mathbf{d}\left(\mathrm{xy}\right)\mathbf{=}\mathbf{x}\text{ }\mathbf{dy}\mathbf{+}\mathbf{y}\text{ }\mathbf{dx}$.

Given that,$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\frac{x-y-1}{x+y+5}\text{ }\mathrm{......}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Evaluate the equation (1).

$\begin{array}{c}\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\frac{x-y-1}{x+y+5}\\ \left(x+y+5\right)\mathbf{dy}\mathbf{=}\left(x-y-1\right)\mathbf{dx}\\ \mathbf{x}\text{ }\mathbf{dy}\mathbf{+}\mathbf{y}\text{ }\mathbf{dy}\mathbf{+}\mathbf{5}\text{ }\mathbf{dy}\mathbf{=}\mathbf{x}\text{ }\mathbf{dx}\mathbf{-}\mathbf{y}\text{ }\mathbf{dx}\mathbf{-}\mathbf{dx}\\ \left(\mathbf{x}\text{ }\mathbf{dy}\mathbf{+}\mathbf{y}\text{ }\mathbf{dx}\right)\mathbf{+}\mathbf{5}\text{ }\mathbf{dy}\mathbf{=}\mathbf{x}\text{ }\mathbf{dx}\mathbf{-}\mathbf{y}\text{ }\mathbf{dy}\mathbf{-}\mathbf{dx}â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots \left(2\right)\end{array}$

Now integrate the equation (2) on both sides.

$∫\left(\mathbf{x}\text{ }\mathbf{dy}\mathbf{+}\mathbf{y}\text{ }\mathbf{dx}\right)\mathbf{+}\mathbf{5}∫\mathbf{dy}\mathbf{=}∫\mathbf{x}\text{ }\mathbf{dx}\mathbf{-}∫\mathbf{y}\text{ }\mathbf{dy}\mathbf{-}∫\mathbf{dx}$

Use the product rule.

$\begin{array}{c}∫\mathbf{d}\left(\mathrm{xy}\right)\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{=}\frac{{\mathbf{x}}^2}{2}\mathbf{-}\frac{{\mathbf{y}}^2}{2}\mathbf{-}\mathbf{x}\mathbf{+}\mathbf{C}\\ \mathbf{xy}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{=}\frac{{\mathbf{x}}^2}{2}\mathbf{-}\frac{{\mathbf{y}}^2}{2}\mathbf{-}\mathbf{x}\mathbf{+}\mathbf{C}\\ \mathbf{xy}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{-}\frac{{\mathbf{x}}^2}{2}\mathbf{+}\frac{{\mathbf{y}}^2}{2}\mathbf{+}\mathbf{x}\mathbf{=}\mathbf{C}\end{array}$

Hence, the solution of the given initial value problem is $\mathbf{xy}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{-}\frac{{\mathbf{x}}^2}{2}\mathbf{+}\frac{{\mathbf{y}}^2}{2}\mathbf{+}\mathbf{x}\mathbf{=}\mathbf{C}$.