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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

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# (a) Show that ${\varphi }\left(\mathbf{x}\right){\mathbf{=}}{{\mathbf{x}}}^{{\mathbf{2}}}$ is an explicit solution to ${\mathbf{x}}\frac{\mathbf{dy}}{\mathbf{dx}}{\mathbf{=}}{\mathbf{2}}{\mathbf{y}}$ on the interval $\left(-\infty ,\infty \right)$.(b) Show that ${\varphi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{{\mathbf{e}}}^{{\mathbf{x}}}{\mathbf{-}}{\mathbf{x}}$, is an explicit solution to $\frac{\mathbf{dy}}{\mathbf{dx}}{\mathbf{+}}{{\mathbf{y}}}^{{\mathbf{2}}}{\mathbf{=}}{{\mathbf{e}}}^{{\mathbf{x}}}{\mathbf{+}}\left(\mathbf{1}\mathbf{-}\mathbf{2}\mathbf{x}\right){{\mathbf{e}}}^{{\mathbf{x}}}{\mathbf{+}}{{\mathbf{x}}}^{{\mathbf{2}}}{\mathbf{-}}{\mathbf{1}}$ on the interval $\left(-\infty ,\infty \right)$.(c) Show that ${\varphi }\left(\mathbf{x}\right){\mathbf{=}}{{\mathbf{x}}}^{{\mathbf{2}}}{\mathbf{-}}{{\mathbf{x}}}^{\mathbf{-}\mathbf{1}}$ is an explicit solution to ${{\mathbf{x}}}^{{\mathbf{2}}}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}{\mathbf{=}}{\mathbf{2}}{\mathbf{y}}$ on the interval $\left(0,\infty \right)$.

1. Proved
2. Proved
3. Proved
See the step by step solution

## Step 1(a): Showing that ϕx=x2 is an explicit solution to xdydx=2y

Firstly, differentiate the given function concerning x.

Therefore, $\mathrm{\varphi }\text{'}\left(\mathrm{x}\right)=2\mathrm{x},$ for all x on the interval $\left(-\infty ,\infty \right)$,

Substituting the value of for y in the given equation,

$\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}=2\mathrm{y}\phantom{\rule{0ex}{0ex}}\mathrm{x}×\mathrm{\varphi }\text{'}\left(\mathrm{x}\right)=2×\mathrm{\varphi }\left(\mathrm{x}\right)\phantom{\rule{0ex}{0ex}}\mathrm{x}×\left(2\mathrm{x}\right)=2×\left({\mathrm{x}}^{2}\right)\phantom{\rule{0ex}{0ex}}2{\mathrm{x}}^{2}=2{\mathrm{x}}^{2}$

This means, the function $\varphi \left(\mathrm{x}\right)$, substituted in the given equation, satisfies the equation for all x in the interval $\left(-\infty ,\infty \right)$.

Hence, ${\varphi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{{\mathbf{x}}}^{{\mathbf{2}}}$ is an explicit solution to the equation ${\mathbf{x}}\frac{\mathbf{dy}}{\mathbf{dx}}{\mathbf{=}}{\mathbf{2}}{\mathbf{y}}$, for all x on the interval $\left(-\infty ,\infty \right)$.

## Step 2(b): Show that ϕ(x)=ex-x, is an explicit solution to dydx+y2=ex+(1-2x)ex+x2-1a2+b2

Firstly, differentiate the given function concerning x.

Therefore, $\varphi \text{'}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}}-1$, for all x on the interval $\left(-\infty ,\infty \right)$.

Substituting the value of $\varphi \left(\mathrm{x}\right)$ for y in the L.H.S. (Left-hand side) of the given equation,

Which is the same as the R.H.S. (Right-hand side) of the given equation,

Hence, ${\mathbit{\varphi }}\left(x\right){\mathbf{=}}{{\mathbf{e}}}^{{\mathbf{x}}}{\mathbf{-}}{\mathbf{x}}$ is an explicit solution to the equation $\frac{\mathbf{dy}}{\mathbf{dx}}{\mathbf{+}}{{\mathbf{y}}}^{{\mathbf{2}}}{\mathbf{=}}{{\mathbf{e}}}^{{\mathbf{x}}}{\mathbf{+}}\left(\mathbf{1}\mathbf{-}\mathbf{2}\mathbf{x}\right){{\mathbf{e}}}^{{\mathbf{x}}}{\mathbf{+}}{{\mathbf{x}}}^{{\mathbf{2}}}{\mathbf{-}}{\mathbf{1}}$, for all x on the interval $\left(-\infty ,\infty \right)$.

## Step 3(c): Show that ϕx=x2-x-1 is an explicit solution to x2d2ydx2=2y

Firstly, we will differentiate the given function concerning x.

Therefore, $\varphi \text{'}\left(\mathrm{x}\right)=2\mathrm{x}-\left(-1\right)×{\mathrm{x}}^{-2}=2\mathrm{x}+{\mathrm{x}}^{-2}$, for all x in the interval $\left(0,\infty \right)$

And $\varphi \text{'}\text{'}\left(\mathrm{x}\right)=2+\left(-2\right){\mathrm{x}}^{-3}=2-2{\mathrm{x}}^{-3}$, for all x in the interval $\left(0,\infty \right)$

Substituting the value of for y in the given equation,

${\mathrm{x}}^{2}\frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}=2\mathrm{y}\phantom{\rule{0ex}{0ex}}{\mathrm{x}}^{2}×\mathrm{\varphi }\text{'}\text{'}\left(\mathrm{x}\right)=2×\mathrm{\varphi }\left(\mathrm{x}\right)\phantom{\rule{0ex}{0ex}}{\mathrm{x}}^{2}×\left(2-2{\mathrm{x}}^{-3}\right)=2×\left({\mathrm{x}}^{2}-{\mathrm{x}}^{-1}\right)\phantom{\rule{0ex}{0ex}}2{\mathrm{x}}^{2}-2{\mathrm{x}}^{-1}=2{\mathrm{x}}^{2}-2{\mathrm{x}}^{-1}$

This means, the function $\mathrm{\varphi }\left(\mathrm{x}\right)$, substituted in the given equation, satisfies the equation, for all x in the interval $\left(0,\infty \right)$.

So, ${\varphi }\left(\mathbf{x}\right){\mathbf{=}}{{\mathbf{x}}}^{{\mathbf{2}}}{\mathbf{-}}{{\mathbf{x}}}^{\mathbf{-}\mathbf{1}}$ is an explicit solution to the equation ${{\mathbf{x}}}^{{\mathbf{2}}}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}{\mathbf{=}}{\mathbf{2}}{\mathbf{y}}$, for all x in the interval $\mathbf{\left(}\mathbf{0}\mathbf{,}{\infty }{\mathbf{\right)}}$.

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