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Q2.2-10E

Expert-verified
Found in: Page 46

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

In problem 7-16, solve the equation. $\frac{\mathbf{dy}}{\mathbf{dx}}{\mathbf{=}}\frac{\mathbf{x}}{{\mathbf{y}}^{\mathbf{2}}\sqrt{\mathbf{1}\mathbf{+}\mathbf{x}}}$

The solution of the given differential equation is ${{\mathbf{y}}}^{3}{\mathbf{=}}{\mathbf{2}}\left[\mathbf{3}\mathbf{x}\sqrt{1+x}\mathbf{-}\mathbf{2}{\left(1+x\right)}^{\frac{3}{2}}\right]{\mathbf{+}}{\mathbf{C}}$.

See the step by step solution

Step 1: Concept of Separable Differential Equation

A first-order ordinary differential equation $\frac{\mathrm{dy}}{\mathrm{dx}}{\mathbf{=}}{\mathbf{f}}\left(x,y\right)$ is referred to as separable if the function in the right-hand side of the equation is expressed as a product of two functions ${\mathbf{g}}\left(x\right)$that is a function of${\mathbf{x}}$alone and${\mathbf{h}}\left(y\right)$that is a function of${\mathbf{y}}$ alone.

A separable differential equation can be expressed as $\frac{\mathrm{dy}}{\mathrm{dx}}{\mathbf{=}}{\mathbf{g}}\left(x\right){\text{ }}{\mathbf{h}}\left(y\right)$. By separating the variables, the equation follows $\frac{\mathrm{dy}}{\mathbf{h}\left(y\right)}{\mathbf{=}}{\mathbf{g}}\left(x\right){\mathbf{dx}}$. Then, on direct integration of both sides, the solution of the differential equation is determined.

Step 2: Solution of the Equation

The given equation is

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}}{{\mathrm{y}}^{2}\sqrt{1+\mathrm{x}}}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

After separating the variables, equation (1) can be written as

${\mathrm{y}}^{2}\text{ }\mathrm{dy}=\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}}}\mathrm{dx}\cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)$

Integrate both sides of equation (2). It results,

$\begin{array}{c}\int {\mathrm{y}}^{2}\text{ }\mathrm{dy}=\int \frac{\mathrm{x}}{\sqrt{1+\mathrm{x}}}\mathrm{dx}\\ \int {\mathrm{y}}^{2}\text{ }\mathrm{dy}=2\int \frac{\mathrm{x}}{2\sqrt{1+\mathrm{x}}}\mathrm{dx}\\ \int {\mathrm{y}}^{2}\text{ }\mathrm{dy}=2\int \mathrm{x}\text{ }\mathrm{d}\left(\sqrt{1+\mathrm{x}}\right)\\ \int {\mathrm{y}}^{2}\text{ }\mathrm{dy}=2\left[\mathrm{x}\sqrt{1+\mathrm{x}}-\int \sqrt{1+\mathrm{x}}\text{ }\mathrm{dx}\right]\text{ }\left[\int \mathrm{u}\text{ }\mathrm{dv}=\mathrm{uv}-\int \mathrm{v}\text{ }\mathrm{du}\right]\end{array}$

$\begin{array}{c}\frac{{\mathrm{y}}^{3}}{3}=2\left[\mathrm{x}\sqrt{1+\mathrm{x}}-\frac{2}{3}{\left(1+\mathrm{x}\right)}^{\frac{3}{2}}\right]+\mathrm{k}\left[\mathrm{k}=\mathrm{IntegratingConstant}\right]\\ {\mathrm{y}}^{3}=2\left[3\mathrm{x}\sqrt{1+\mathrm{x}}-2{\left(1+\mathrm{x}\right)}^{\frac{3}{2}}\right]+3\mathrm{k}\\ {\mathrm{y}}^{3}=2\left[3\mathrm{x}\sqrt{1+\mathrm{x}}-2{\left(1+\mathrm{x}\right)}^{\frac{3}{2}}\right]+\mathrm{C}\text{ }\left[\mathrm{C}=3\mathrm{k}=\mathrm{Constant}\right]\end{array}$

Therefore, the solution of the given equation is ${{\mathbf{y}}}^{3}{\mathbf{=}}{\mathbf{2}}\left[\mathbf{3}\mathbf{x}\sqrt{1+x}\mathbf{-}\mathbf{2}{\left(1+x\right)}^{\frac{3}{2}}\right]{\mathbf{+}}{\mathbf{C}}$.