91Ó°ÊÓ

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 ${\mathbf{x}}_0∈\mathbf{I}$ 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:$∫\mathbf{udv}\mathbf{=}\mathbf{uv}\mathbf{-}∫\mathbf{vdu}$.
• $∫{\mathbf{x}}^a\mathbf{dx}\mathbf{=}\frac{{\mathbf{x}}^{a+1}}{a+1}\mathbf{+}\mathbf{C}$.
• $∫\frac{1}{x}\mathbf{dx}\mathbf{=}\mathbf{ln}\left|x\right|\mathbf{+}\mathbf{C}$
• $\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{+}\mathbf{P}\left(x\right)\mathbf{y}\mathbf{=}\mathbf{Q}\left(x\right)$

Given that, $\frac{\mathrm{dx}}{\mathrm{dt}}\mathbf{-}\frac{x}{t-1}\mathbf{=}{\mathbf{t}}^2\mathbf{+}\mathbf{2}\text{ }\mathrm{......}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Let $\mathbf{P}\left(t\right)\mathbf{=}\mathbf{-}\frac{1}{t-1}$.

Find the value of $\mathrm{μ}\left(t\right)$.

$\begin{array}{c}\mathrm{μ}\left(t\right)\mathbf{=}{\mathbf{e}}^{∫\mathbf{P}\left(t\right)\mathbf{dt}}\\ \mathbf{=}{\mathbf{e}}^{\mathbf{-}∫\frac{1}{t-1}\mathbf{dt}}\\ \mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{ln}\left|t-1\right|}\\ \mathbf{=}\frac{1}{t-1}\end{array}$

Multiply $\frac{1}{t-1}$ in equation (1) on both sides.

$\begin{array}{c}\frac{1}{t-1}\frac{\mathrm{dx}}{\mathrm{dt}}\mathbf{-}\frac{x}{{\left(t-1\right)}^2}\mathbf{=}\frac{{\mathbf{t}}^2\mathbf{+}\mathbf{2}}{t-1}\\ \frac{d}{\mathrm{dt}}\left[\frac{1}{t-1}\mathbf{x}\right]\mathbf{=}\frac{{\mathbf{t}}^2\mathbf{+}\mathbf{2}}{t-1}\end{array}$

Now integrate the equation on both sides.

$\begin{array}{c}∫\frac{d}{\mathrm{dt}}\left[\frac{1}{t-1}\mathbf{x}\right]\mathbf{dt}\mathbf{=}∫\frac{{\mathbf{t}}^2\mathbf{+}\mathbf{2}}{t-1}\mathbf{dt}\\ \frac{x}{t-1}\mathbf{=}∫\frac{{\mathbf{t}}^2\mathbf{+}\mathbf{2}}{t-1}\mathbf{dt}\text{ }\mathrm{......}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\end{array}$

Find the value of $∫\frac{{\mathbf{t}}^2\mathbf{+}\mathbf{2}}{t-1}\mathbf{dt}$separately.

Let us take $\mathbf{u}\mathbf{=}\mathbf{t}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{t}\mathbf{=}\mathbf{u}\mathbf{+}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{dt}\mathbf{=}\mathbf{du}$.

Use the integration by parts formula.

$\begin{array}{c}∫\frac{{\mathbf{t}}^2\mathbf{+}\mathbf{2}}{t-1}\mathbf{dt}\mathbf{=}∫\frac{{\left(u+1\right)}^2\mathbf{+}\mathbf{2}}{u}\mathbf{du}\\ \mathbf{=}∫\frac{{\mathbf{u}}^2\mathbf{+}\mathbf{2}\mathbf{u}\mathbf{+}\mathbf{1}\mathbf{+}\mathbf{2}}{u}\mathbf{du}\\ \mathbf{=}∫\left(\mathbf{u}\mathbf{+}\mathbf{2}\mathbf{+}\frac{3}{u}\right)\mathbf{du}\\ \mathbf{=}\frac{{\mathbf{u}}^2}{2}\mathbf{+}\mathbf{2}\mathbf{u}\mathbf{+}\mathbf{3}\mathbf{ln}\left|u\right|\mathbf{+}{\mathbf{C}}_1\\ \mathbf{=}\frac{{\mathbf{t}}^2\mathbf{-}\mathbf{2}\mathbf{t}\mathbf{+}\mathbf{1}}{2}\mathbf{+}\mathbf{2}\mathbf{t}\mathbf{-}\mathbf{2}\mathbf{+}\mathbf{3}\mathbf{ln}\left|t-1\right|\mathbf{+}{\mathbf{C}}_1\\ \mathbf{=}\frac{{\mathbf{t}}^2}{2}\mathbf{-}\mathbf{t}\mathbf{+}\frac{1}{2}\mathbf{+}\mathbf{2}\mathbf{t}\mathbf{-}\mathbf{2}\mathbf{+}\mathbf{3}\mathbf{ln}\left|t-1\right|\mathbf{+}{\mathbf{C}}_1\\ \mathbf{=}\frac{{\mathbf{t}}^2}{2}\mathbf{+}\mathbf{t}\mathbf{+}\mathbf{3}\mathbf{ln}\left|t-1\right|\mathbf{-}\frac{3}{2}\mathbf{+}{\mathbf{C}}_1\end{array}$

Now substitute in equation (2)

$\begin{array}{c}\frac{x}{t-1}\mathbf{=}\frac{{\mathbf{t}}^2}{2}\mathbf{+}\mathbf{t}\mathbf{+}\mathbf{3}\mathbf{ln}\left|t-1\right|\mathbf{-}\frac{3}{2}\mathbf{+}{\mathbf{C}}_1\\ \mathbf{x}\mathbf{=}\frac{\left(t-1\right){\mathbf{t}}^2}{2}\mathbf{+}\mathbf{t}\left(t-1\right)\mathbf{+}\mathbf{3}\left(t-1\right)\mathbf{ln}\left|t-1\right|\mathbf{-}\frac{3}{2}\left(t-1\right)\mathbf{+}\mathbf{C}\left(t-1\right)\end{array}$

Hence, the solution of the given initial value problem is

$\mathbf{x}\mathbf{=}\frac{\left(t-1\right){\mathbf{t}}^2}{2}\mathbf{+}\mathbf{t}\left(t-1\right)\mathbf{+}\mathbf{3}\left(t-1\right)\mathbf{ln}\left|t-1\right|\mathbf{-}\frac{3}{2}\left(t-1\right)\mathbf{+}\mathbf{C}\left(t-1\right)$