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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.

Homogeneous:it is satisfying only the function turns into$\frac{y}{x}$ format.

Linear coefficient:here we need to convert both the variables into another two variables. For example, x as u and y as v.

Formulae to be used:

• Integration by parts:. $鈭�\mathbf{udv}\mathbf{=}\mathbf{uv}\mathbf{-}鈭�\mathbf{vdu}$
• $鈭�{\mathbf{x}}^a\mathbf{dx}\mathbf{=}\frac{{\mathbf{x}}^{a+1}}{a+1}\mathbf{+}\mathbf{C}$
  
• $鈭�\frac{1}{x}\mathbf{dx}\mathbf{=}\mathbf{ln}\left|x\right|\mathbf{+}\mathbf{C}$
  

Given that,$\left(2x-2y-8\right)\mathbf{dx}\mathbf{+}\left(x-3y-6\right)\mathbf{dy}\mathbf{=}\mathbf{0}\text{鈥�}\mathrm{......}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Since the given equation is linear coefficients. So, we can take

$\begin{array}{l}\mathbf{x}\mathbf{=}\mathbf{u}\mathbf{+}\mathbf{h}鈫�\mathbf{dx}\mathbf{=}\mathbf{du}\\ \mathbf{y}\mathbf{=}\mathbf{v}\mathbf{+}\mathbf{k}鈫�\mathbf{dy}\mathbf{=}\mathbf{dv}\end{array}$

Evaluate the equation (1).

$\begin{array}{c}\left(2x-2y-8\right)\mathbf{dx}\mathbf{+}\left(x-3y-6\right)\mathbf{dy}\mathbf{=}\mathbf{0}\\ \frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{-}\frac{2x-2y-8}{x-3y-6}\\ \frac{\mathrm{dv}}{\mathrm{du}}\mathbf{=}\mathbf{-}\frac{\mathbf{2}\left(u+h\right)\mathbf{-}\mathbf{8}\mathbf{-}\mathbf{2}\left(v+k\right)}{u+h-3v-3k-6}\\ \mathbf{=}\mathbf{-}\frac{2u-2v+2h-2k-8}{u-3v+h-3k-6}\end{array}$

Using the condition, we get,

$\begin{array}{c}2h-2k-8=0\\ h-3k-6=0\end{array}$

Solve the above equations to find the values ofhandk.

role="math" localid="1663953044785" $\begin{array}{c}4k+4=0\\ 4k=-4\\ k=-1\end{array}$

Substitute k = -1.

$\begin{array}{c}h+3-6=0\\ h=3\end{array}$

Then,

$\begin{array}{c}\frac{\mathrm{dv}}{\mathrm{du}}\mathbf{=}鈭�\frac{\mathbf{2}\mathbf{u}\mathbf{-}2\mathbf{v}}{\mathbf{u}鈭�3\mathbf{v}}\\ \mathbf{=}\frac{\mathbf{u}\left(\mathbf{2}\mathbf{-}2\frac{v}{u}\right)}{\mathbf{u}\left(3\frac{v}{u}鈭�1\right)}\\ \mathbf{=}\frac{\mathbf{2}\mathbf{-}2\frac{v}{u}}{3\frac{v}{u}鈭�1}\end{array}$

So, the founded equation is homogeneous.

Let us take $\mathbf{s}\mathbf{=}\frac{v}{u}鈫�\frac{\mathrm{dv}}{\mathrm{du}}\mathbf{=}\mathbf{s}\mathbf{+}\mathbf{u}\frac{\mathrm{ds}}{\mathrm{du}}$.

Then,

$\begin{array}{c}\mathbf{s}\mathbf{+}\mathbf{u}\frac{\mathrm{ds}}{\mathrm{du}}\mathbf{=}\frac{\mathbf{2}\mathbf{-}2\mathbf{s}}{3s鈭�1}\\ \mathbf{u}\frac{\mathrm{ds}}{\mathrm{du}}\mathbf{=}\frac{\mathbf{2}\mathbf{-}2\mathbf{s}}{3\mathbf{s}鈭�1}\mathbf{-}\mathbf{s}\\ \mathbf{=}\frac{\mathbf{2}\mathbf{-}2\mathbf{s}+\mathbf{s}\mathbf{-}3{\mathbf{s}}^2}{3\mathbf{s}鈭�1}\\ \mathbf{=}\frac{\mathbf{2}\mathbf{-}\mathbf{s}\mathbf{-}3{\mathbf{s}}^2}{3\mathbf{s}鈭�1}\\ \frac{3\mathbf{s}鈭�1}{3{\mathbf{s}}^2\mathbf{+}\mathbf{s}\mathbf{-}\mathbf{2}}\mathbf{ds}\mathbf{=}\mathbf{-}\frac{\mathrm{du}}{u}鈰�鈰�鈰�鈰�鈰�鈰�\left(2\right)\end{array}$

Now integrate the equation (2) on both sides.

$\begin{array}{c}鈭�\frac{1鈭�3s}{\mathbf{-}\mathbf{3}{\mathbf{s}}^2\mathbf{-}\mathbf{s}\mathbf{+}\mathbf{2}}\mathbf{ds}\mathbf{=}鈭�\frac{\mathrm{du}}{u}\\ \frac{1}{5}\mathbf{ln}\left|\mathbf{3}\mathbf{s}鈭�2\right|+\frac{4}{5}\mathbf{ln}\left|1+s\right|\mathbf{=}\mathbf{ln}\left|u\right|\mathbf{+}{\mathbf{C}}_1\\ \mathbf{ln}\left|\mathbf{3}\mathbf{s}鈭�2\right|\mathbf{+}\mathbf{4}\mathbf{ln}\left|1+s\right|\mathbf{=}\mathbf{5}\mathbf{ln}\left|u\right|\mathbf{+}{\mathbf{C}}_2\\ \mathbf{ln}\left|\left(\mathbf{3}\mathbf{s}鈭�2\right){\left(1+s\right)}^4\right|\mathbf{=}\mathbf{ln}\left|{\mathbf{u}}^5\right|\mathbf{+}{\mathbf{C}}_2\\ \left(\mathbf{3}\mathbf{s}鈭�2\right){\left(1+s\right)}^4{\mathbf{u}}^5\mathbf{=}{\mathbf{C}}_3\end{array}$

Again, substitute the value of s, u and v

$\begin{array}{c}\left(\mathbf{3}\left(\frac{v}{u}\right)鈭�2\right){\left(\mathbf{1}\mathbf{+}\left(\frac{v}{u}\right)\right)}^4{\mathbf{u}}^5\mathbf{=}{\mathbf{C}}_3\\ \left(\frac{\mathbf{3}\mathbf{v}鈭�2u}{u}\right)\left(\frac{{\left(u+v\right)}^4}{{\mathbf{u}}^4}\right){\mathbf{u}}^5\mathbf{=}{\mathbf{C}}_3\\ \left(3v鈭�2u\right){\left(u+v\right)}^4\mathbf{=}{\mathbf{C}}_3\\ (3y+3鈭�2x+6){\left(x-3+y+1\right)}^4=C\\ (3y鈭�2x+9){\left(x+y-2\right)}^4\mathbf{=}\mathbf{C}\end{array}$

Hence, the solution of the given initial value problem is

$\left(\mathbf{3}\mathbf{y}鈭�2x+\mathbf{9}\right){\left(x+y-2\right)}^4\mathbf{=}\mathbf{C}$.