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### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# Question: In Problems 1-30, solve the equation.$\left(3x-y-5\right){\mathbf{dx}}{\mathbf{+}}\left(x-y+1\right){\mathbf{dy}}{\mathbf{=}}{\mathbf{0}}$

$\left[{\left(y-4\right)}^{2}\mathbf{-}\mathbf{3}{\left(x-3\right)}^{2}\right]{\left|\frac{\sqrt{3}\left(x-3\right)\mathbf{+}\left(y-4\right)}{\sqrt{3}\left(x-3\right)\mathbf{-}\left(y-4\right)}\right|}^{\frac{1}{\sqrt{3}}}\mathbf{=}\mathbf{C}$

See the step by step solution

## Step 1: Definition and concepts to be used

Definition of Initial Value Problem: By an initial value problem for an nth-order differential equation $\mathbf{F}\left(\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{,}...\mathbf{,}\frac{{\mathbf{d}}^{n}\mathbf{y}}{{\mathbf{dx}}^{n}}\right)\mathbf{=}\mathbf{0}$ 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 ${\mathbf{x}}_{0}\in \mathbf{I}$ and ${\mathbf{y}}_{0}\mathbf{,}{\mathbf{y}}_{1}\mathbf{,}...\mathbf{,}{\mathbf{y}}_{n-1}$ 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: $\int \mathbf{udv}\mathbf{=}\mathbf{uv}\mathbf{-}\int \mathbf{vdu}$.
• $\int {\mathbf{x}}^{a}\mathbf{dx}\mathbf{=}\frac{{\mathbf{x}}^{a+1}}{a+1}\mathbf{+}\mathbf{C}$.
• $\int \frac{1}{x}\mathbf{dx}\mathbf{=}\mathbf{ln}\left|x\right|\mathbf{+}\mathbf{C}$.

## Step 2: Given information and simplification

Given that, $\left(3x-y-5\right)\mathbf{dx}\mathbf{+}\left(x-y+1\right)\mathbf{dy}\mathbf{=}\mathbf{0}\text{\hspace{0.17em}}......\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

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

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

Evaluate the equation (1).

$\begin{array}{c}\left(3x-y-5\right)\mathbf{dx}\mathbf{+}\left(x-y+1\right)\mathbf{dy}\mathbf{=}\mathbf{0}\\ \frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\frac{3x-y-5}{x-y+1}\\ \frac{\mathrm{dv}}{\mathrm{du}}\mathbf{=}\frac{\mathbf{3}\left(u+h\right)\mathbf{-}\left(v+k\right)\mathbf{-}\mathbf{5}}{u+h-v-k+1}\\ \mathbf{=}\frac{3u-v+3h-k-5}{u-v+h-k+1}\end{array}$

Using the condition, we get,

$\begin{array}{c}3h-k-5=0\\ h-k+1=0\end{array}$

Solve the above equations to find the values of h and k.

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

Substitute h = 3.

$\begin{array}{c}3-k+1=0\\ k=4\end{array}$

Then,

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

So, the founded equation is homogeneous.

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

Then,

$\begin{array}{c}\mathbf{s}\mathbf{+}\mathbf{u}\frac{\mathrm{ds}}{\mathrm{du}}\mathbf{=}\frac{3-s}{1-s}\\ \mathbf{u}\frac{\mathrm{ds}}{\mathrm{du}}\mathbf{=}\frac{3-s}{1-s}\mathbf{-}\mathbf{s}\\ \mathbf{=}\frac{\mathbf{3}\mathbf{-}\mathbf{s}\mathbf{-}\mathbf{s}\mathbf{+}{\mathbf{s}}^{2}}{1-s}\\ \mathbf{=}\frac{\mathbf{3}\mathbf{-}\mathbf{2}\mathbf{s}\mathbf{+}{\mathbf{s}}^{2}}{1-s}\\ \frac{1-s}{{\mathbf{s}}^{2}\mathbf{-}\mathbf{2}\mathbf{s}\mathbf{+}\mathbf{3}}\mathbf{ds}\mathbf{=}\frac{\mathrm{du}}{u}\end{array}$

## Step 3: Evaluation method

Now integrate the equation (2) on both sides.

$\int \frac{1-s}{{\mathbf{s}}^{2}\mathbf{-}\mathbf{2}\mathbf{s}\mathbf{+}\mathbf{3}}\mathbf{ds}\mathbf{=}\int \frac{\mathrm{du}}{u}$

Find the value of $\int \frac{1-s}{{\mathbf{s}}^{2}\mathbf{-}\mathbf{2}\mathbf{s}\mathbf{+}\mathbf{3}}\mathbf{ds}$ separately.

Let $\mathbf{t}\mathbf{=}{\mathbf{s}}^{2}\mathbf{-}\mathbf{2}\mathbf{s}\mathbf{+}\mathbf{3}\to \mathbf{dt}\mathbf{=}\mathbf{-}\mathbf{2}\left(1-s\right)\mathbf{ds}$.

Then,

$\begin{array}{c}\int \frac{1-s}{{\mathbf{s}}^{2}\mathbf{-}\mathbf{2}\mathbf{s}\mathbf{+}\mathbf{3}}\mathbf{ds}\mathbf{=}\mathbf{-}\int \frac{1}{t}\frac{\mathrm{dt}}{2}\\ \mathbf{=}\mathbf{-}\frac{1}{2}\int \frac{1}{t}\mathbf{dt}\\ \mathbf{=}\mathbf{-}\frac{1}{2}\mathbf{ln}\left|t\right|\\ \mathbf{=}\mathbf{-}\frac{1}{2}\mathbf{ln}\left|{\mathbf{s}}^{2}\mathbf{-}\mathbf{2}\mathbf{s}\mathbf{+}\mathbf{3}\right|\end{array}$

Substitute the value here.

$\begin{array}{c}\int \frac{1-s}{{\mathbf{s}}^{2}\mathbf{-}\mathbf{2}\mathbf{s}\mathbf{+}\mathbf{3}}\mathbf{ds}\mathbf{=}\int \frac{\mathrm{du}}{u}\\ \mathbf{-}\frac{1}{2}\mathbf{ln}\left|{\mathbf{s}}^{2}\mathbf{-}\mathbf{2}\mathbf{s}\mathbf{+}\mathbf{3}\right|\mathbf{=}\mathbf{ln}\left|u\right|\mathbf{+}{\mathbf{C}}_{1}\\ \mathbf{ln}\left|{\mathbf{s}}^{2}\mathbf{-}\mathbf{2}\mathbf{s}\mathbf{+}\mathbf{3}\right|\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{ln}\left|u\right|\mathbf{+}{\mathbf{C}}_{2}\\ \mathbf{=}\mathbf{-}\mathbf{ln}\left|{\mathbf{u}}^{2}\right|\mathbf{+}{\mathbf{C}}_{2}\\ \left({\mathbf{s}}^{2}\mathbf{-}\mathbf{2}\mathbf{s}\mathbf{+}\mathbf{3}\right){\mathbf{u}}^{2}\mathbf{=}{\mathbf{C}}_{3}\end{array}$

Again, substitute the value of s, u and v

$\begin{array}{c}\left({\left(\frac{v}{u}\right)}^{2}-\mathbf{2}\left(\frac{v}{u}\right)+3\right){\mathbf{u}}^{2}\mathbf{=}{\mathbf{C}}_{2}\\ \left(\frac{{\mathbf{v}}^{2}}{{\mathbf{u}}^{2}}-\mathbf{2}\frac{v}{u}+3\right){\mathbf{u}}^{2}\mathbf{=}{\mathbf{C}}_{2}\\ \frac{{\mathbf{v}}^{2}-\mathbf{2}\mathbf{uv}+3{\mathbf{u}}^{2}}{{\mathbf{u}}^{2}}{\mathbf{u}}^{2}\mathbf{=}{\mathbf{C}}_{2}\\ {\mathbf{v}}^{2}-\mathbf{2}\mathbf{uv}+3{\mathbf{u}}^{2}\mathbf{=}{\mathbf{C}}_{2}\end{array}$

$\left[{\left(y-4\right)}^{2}\mathbf{-}\mathbf{3}{\left(x-3\right)}^{2}\right]{\left|\frac{\sqrt{3}\left(x-3\right)\mathbf{+}\left(y-4\right)}{\sqrt{3}\left(x-3\right)\mathbf{-}\left(y-4\right)}\right|}^{\frac{1}{\sqrt{3}}}\mathbf{=}\mathbf{C}$

Hence, the solution of given initial value problem is .

$\left[{\left(y-4\right)}^{2}\mathbf{-}\mathbf{3}{\left(x-3\right)}^{2}\right]{\left|\frac{\sqrt{3}\left(x-3\right)\mathbf{+}\left(y-4\right)}{\sqrt{3}\left(x-3\right)\mathbf{-}\left(y-4\right)}\right|}^{\frac{1}{\sqrt{3}}}{\mathbf{=}}{\mathbf{C}}$