91Ó°ÊÓ

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

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

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

Formulae to be used:

• Integration by parts: role="math" localid="1664255077420" $\mathbf{∫}\mathit{u}\mathit{d}\mathit{v}\mathbf{=}\mathit{u}\mathit{v}\mathbf{-∫}\mathit{v}\mathit{d}\mathit{u}$.
• role="math" localid="1664255091624" $∫{\mathbf{x}}^{\mathbf{a}}\mathbf{dx}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{a}\mathbf{+}\mathbf{1}}}{\mathbf{a}\mathbf{+}\mathbf{1}}\mathbf{+}\mathbf{C}$
  
• role="math" localid="1664255105362" $∫\frac{\mathbf{1}}{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}}\mathbf{dx}\mathbf{=}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{x}\mathbf{+}\mathbf{C}$
  

Given that, $\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{1}{1+{\mathrm{t}}^2}-\mathrm{y},      \mathrm{y}\left(2\right)=3          ......\left(1\right)$

Evaluate the equation (1).

$\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{1}{1+{\mathrm{t}}^2}-\mathrm{y}$

Let $\mathrm{P}=1$.

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

$$\begin{gathered} \mathrm{μ}\left(\mathrm{t}\right)={\mathrm{e}}^{∫\mathrm{P}\left(\mathrm{t}\right)\mathrm{dt}} \\ ={\mathrm{e}}^{∫\mathrm{dt}} \\ ={\mathrm{e}}^{\mathrm{t}} \end{gathered}$$

Multiply ${\mathrm{e}}^{\mathrm{t}}$on both sides.

$$\begin{gathered} {\mathrm{e}}^{\mathrm{t}}\frac{\mathrm{dy}}{\mathrm{dt}}+{\mathrm{e}}^{\mathrm{t}}\mathrm{y}=\frac{{\mathrm{e}}^{\mathrm{t}}}{1+{\mathrm{t}}^2} \\ \frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{e}}^{\mathrm{t}}\mathrm{y}\right]=\frac{{\mathrm{e}}^{\mathrm{t}}}{1+{\mathrm{t}}^2} \end{gathered}$$

Now integrate the equation on both sides.

$$\begin{gathered} ∫\frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{e}}^{\mathrm{t}}\mathrm{y}\right]\mathrm{dx}=∫\frac{{\mathrm{e}}^{\mathrm{t}}}{1+{\mathrm{t}}^2}\mathrm{dt} \\ {\mathrm{e}}^{\mathrm{t}}\mathrm{y}=∫\frac{{\mathrm{e}}^{\mathrm{t}}}{1+{\mathrm{t}}^2}\mathrm{dt}+{\mathrm{C}}_1 \\ \mathrm{y}={\mathrm{e}}^{-\mathrm{t}}∫\frac{{\mathrm{e}}^{\mathrm{t}}}{1+{\mathrm{t}}^2}\mathrm{dt}+{\mathrm{Ce}}^{-\mathrm{t}} \end{gathered}$$

Given that, $\mathrm{y}\left(2\right)=3$

Then, t=2 and y=3.

Substitute the value in equation (3) to get the value of C.

$$\begin{gathered} \mathrm{y}={\mathrm{e}}^{-\mathrm{t}}∫\frac{{\mathrm{e}}^{\mathrm{t}}}{1+{\mathrm{t}}^2}\mathrm{dt}+{\mathrm{Ce}}^{-\mathrm{t}} \\ 3={\mathrm{e}}^{-2}{∫}_0^2\frac{{\mathrm{e}}^{\mathrm{t}}}{1+{\mathrm{t}}^2}\mathrm{dt}+{\mathrm{Ce}}^{-2} \\ 3=0.360424+{\mathrm{Ce}}^{-2} \\ \mathrm{C}={\mathrm{e}}^2\left(3-0.360424\right) \\ ≈19.504 \end{gathered}$$

Substitute the value of C in equation (3).

$$\begin{gathered} \mathrm{y}={\mathrm{e}}^{-3}{∫}_0^3\frac{{\mathrm{e}}^{\mathrm{t}}}{1+{\mathrm{t}}^2}\mathrm{dt}+19.504{\mathrm{e}}^{-3} \\ \mathrm{y}≈1.18834 \end{gathered}$$

Hence, the solution of the given initial value problem is $\mathrm{y}={\mathrm{e}}^{-\mathrm{t}}{∫}_2^{\mathrm{t}}\frac{{\mathrm{e}}^{\mathrm{r}}}{1+{\mathrm{r}}^2}\mathrm{dr}+3{\mathrm{e}}^{-\left(\mathrm{t}-2\right)}$and 1.1883.