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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 ${\mathbf{x}}_0鈭�\mathbf{I}$ 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:$鈭�\mathbf{udv}\mathbf{=}\mathbf{uv}\mathbf{-}鈭�\mathbf{vdu}$.
• data-custom-editor="chemistry" $鈭�{\mathbf{x}}^a\mathbf{dx}\mathbf{=}\frac{{\mathbf{x}}^{a+1}}{a+1}\mathbf{+}\mathbf{C}$.
• $鈭�\frac{1}{{\mathbf{x}}^2\mathbf{+}\mathbf{1}}\mathbf{dx}\mathbf{=}{\mathbf{tan}}^{-1}\mathbf{x}\mathbf{+}\mathbf{C}$

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

Evaluate the equation (1).

Let us assume $\mathbf{t}\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{-}\mathbf{1}$. Differentiate with respect to x.

$\frac{\mathrm{dt}}{\mathrm{dx}}\mathbf{=}\mathbf{2}\mathbf{+}\frac{\mathrm{dy}}{\mathrm{dx}}$

Substitute the values in equation (1)

$\begin{array}{c}\frac{\mathrm{dt}}{\mathrm{dx}}\mathbf{-}\mathbf{2}\mathbf{=}{\mathbf{t}}^2\\ \frac{\mathrm{dt}}{\mathrm{dx}}\mathbf{=}{\mathbf{t}}^2\mathbf{+}\mathbf{2}\\ \frac{1}{{\mathbf{t}}^2\mathbf{+}\mathbf{2}}\mathbf{dt}\mathbf{=}\mathbf{dx}\text{鈥�}\mathrm{......}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\end{array}$

Now integrate the equation (2) on both sides.

$\begin{array}{c}鈭�\frac{1}{{\mathbf{t}}^2\mathbf{+}\mathbf{2}}\mathbf{dt}\mathbf{=}鈭�\mathbf{dx}\\ 鈭�\frac{1}{2\left(\frac{{\mathbf{t}}^2}{2}\mathbf{+}\mathbf{1}\right)}\mathbf{dt}\mathbf{=}鈭�\mathbf{dx}\\ \frac{1}{2}鈭�\frac{1}{\frac{{\mathbf{t}}^2}{2}\mathbf{+}\mathbf{1}}\mathbf{dt}\mathbf{=}鈭�\mathbf{dx}\\ 鈭�\frac{1}{\frac{{\mathbf{t}}^2}{2}\mathbf{+}\mathbf{1}}\mathbf{dt}\mathbf{=}\mathbf{2}鈭�\mathbf{dx}\\ \sqrt{2}{\mathbf{tan}}^{-1}\left(\frac{t}{\sqrt{2}}\right)\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{C}\\ \mathbf{t}\mathbf{=}\sqrt{2}\mathbf{tan}\left(\frac{2x+C}{\sqrt{2}}\right)\end{array}$

Substitute the value of t.

$\begin{array}{c}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{-}\mathbf{1}\mathbf{=}\sqrt{2}\mathbf{tan}\left(\frac{2x+C}{\sqrt{2}}\right)\\ \mathbf{y}\mathbf{=}\sqrt{2}\mathbf{tan}\left(\frac{2x+C}{\sqrt{2}}\right)\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{1}\end{array}$

Hence, the solution of the given initial value problem is $\mathbf{y}\mathbf{=}\sqrt{2}\mathbf{tan}\left(\frac{2x+C}{\sqrt{2}}\right)\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{1}$.