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Found in: Page 79

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.${\text{\hspace{0.17em}}}{{\mathbf{t}}}^{3}{{\mathbf{y}}}^{2}{\mathbf{dt}}{\mathbf{+}}{{\mathbf{t}}}^{4}{{\mathbf{y}}}^{-6}{\mathbf{dy}}{\mathbf{=}}{\mathbf{0}}$

$\mathbf{t}\mathbf{=}{\mathbf{Ce}}^{\left(\frac{1}{\mathbf{7}{\mathbf{y}}^{7}}\right)}$

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.

Formulae to be used:

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

Step 2: Given information and simplification

Given that,${\mathbf{t}}^{3}{\mathbf{y}}^{2}\mathbf{dt}\mathbf{+}{\mathbf{t}}^{4}{\mathbf{y}}^{-6}\mathbf{dy}\mathbf{=}\mathbf{0}\text{\hspace{0.17em}}......\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Evaluate the equation (1).

$\begin{array}{c}{\mathbf{t}}^{3}{\mathbf{y}}^{2}\mathbf{dt}\mathbf{+}{\mathbf{t}}^{4}{\mathbf{y}}^{-6}\mathbf{dy}\mathbf{=}\mathbf{0}\\ {\mathbf{t}}^{3}{\mathbf{y}}^{2}\mathbf{dt}\mathbf{=}\mathbf{-}{\mathbf{t}}^{4}{\mathbf{y}}^{-6}\mathbf{dy}\\ \frac{{\mathbf{t}}^{3}}{{\mathbf{t}}^{4}}\mathbf{dt}\mathbf{=}\mathbf{-}\frac{{\mathbf{y}}^{-6}}{{\mathbf{y}}^{2}}\mathbf{dy}\\ \frac{1}{t}\mathbf{dt}\mathbf{=}\mathbf{-}{\mathbf{y}}^{-8}\mathbf{dy}\text{\hspace{0.17em}}......\mathbf{\left(}\mathbf{2}\mathbf{\right)}\end{array}$

Now integrate the equation (2) on both sides.

$\int \frac{1}{t}\mathbf{dt}\mathbf{=}\mathbf{-}\int {\mathbf{y}}^{-8}\mathbf{dy}......\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

Step 3: Evaluation method

$\begin{array}{c}\int \frac{1}{t}\mathbf{dt}\mathbf{=}\mathbf{-}\int {\mathbf{y}}^{-8}\mathbf{dy}\\ \mathbf{ln}\left|t\right|\mathbf{+}{\mathbf{C}}_{1}\mathbf{=}\frac{1}{7}{\mathbf{y}}^{-7}\\ \mathbf{7}\mathbf{ln}\left|t\right|\mathbf{+}\mathbf{C}\mathbf{=}{\mathbf{y}}^{-7}\\ \mathbf{y}\mathbf{=}{\left(\mathbf{7}\mathbf{ln}\left|t\right|\mathbf{+}\mathbf{C}\right)}^{\mathbf{-}\frac{1}{7}}\end{array}$

Take 7th root on both sides.

$\begin{array}{c}{\mathbf{y}}^{7}\mathbf{=}\frac{1}{\mathbf{7}\mathbf{ln}\left|t\right|\mathbf{+}\mathbf{C}}\\ {\mathbf{y}}^{7}\left(\mathbf{7}\mathbf{ln}\left|t\right|\mathbf{+}\mathbf{C}\right)\mathbf{=}\mathbf{1}\\ \mathbf{t}\mathbf{=}{\mathbf{Ce}}^{\left(\frac{1}{\mathbf{7}{\mathbf{y}}^{7}}\right)}\end{array}$

Hence, the solution of the given initial value problem is ${\mathbf{t}}{\mathbf{=}}{{\mathbf{Ce}}}^{\left(\frac{1}{\mathbf{7}{\mathbf{y}}^{7}}\right)}$.