Q1 E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

(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 $\left(0,\infty \right)$.

Step-by-Step Solution

Answer

1. Proved
2. Proved
3. Proved

Step 1(a): Showing that ϕ x = x 2 is an explicit solution to x dy dx = 2 y

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,

$$\begin{gathered} \mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}=2\mathrm{y} \\ \mathrm{x}\times \mathrm{\varphi }\text{'}\left(\mathrm{x}\right)=2\times \mathrm{\varphi }\left(\mathrm{x}\right) \\ \mathrm{x}\times \left(2\mathrm{x}\right)=2\times \left({\mathrm{x}}^2\right) \\ 2{\mathrm{x}}^2=2{\mathrm{x}}^2 \end{gathered}$$

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 ) = e x - x , is an explicit solution to dy dx + y 2 = e x + ( 1 - 2 x ) e x + x 2 - 1 a 2 + b 2

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,

$$\begin{gathered} \frac{\mathrm{dy}}{\mathrm{dx}}+{\mathrm{y}}^2={\mathrm{e}}^{\mathrm{x}}-1+{\left({\mathrm{e}}^{\mathrm{x}}-\mathrm{x}\right)}^2 \\ \frac{{\displaystyle \mathrm{dy}}}{{\displaystyle \mathrm{dx}}}+{\mathrm{y}}^2={\mathrm{e}}^{\mathrm{x}}-1+{\mathrm{e}}^{2\mathrm{x}}+{\mathrm{x}}^2-2{\mathrm{xe}}^{\mathrm{x}} \\ \frac{{\displaystyle \mathrm{dy}}}{{\displaystyle \mathrm{dx}}}+{\mathrm{y}}^2={\mathrm{e}}^{2\mathrm{x}}+{\mathrm{e}}^{\mathrm{x}}-2{\mathrm{xe}}^{\mathrm{x}}+{\mathrm{x}}^2-1 \\ \frac{{\displaystyle \mathrm{dy}}}{{\displaystyle \mathrm{dx}}}+{\mathrm{y}}^2={\mathrm{e}}^{2\mathrm{x}}+\left(1-2\mathrm{x}\right){\mathrm{e}}^{\mathrm{x}}+{\mathrm{x}}^2-1 \end{gathered}$$

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

Hence, $\mathit{\varphi }\left(x\right)\mathbf{=}{\mathbf{e}}^{\mathbf{x}}\mathbf{-}\mathbf{x}$ 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 ϕ x = x 2 - x - 1 is an explicit solution to x 2 d 2 y dx 2 = 2 y

Firstly, we will differentiate the given function concerning x.

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

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 $(0,\infty )$

Substituting the value of  for y in the given equation,

$$\begin{gathered} {\mathrm{x}}^2\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=2\mathrm{y} \\ {\mathrm{x}}^2\times \mathrm{\varphi }\text{'}\text{'}\left(\mathrm{x}\right)=2\times \mathrm{\varphi }\left(\mathrm{x}\right) \\ {\mathrm{x}}^2\times \left(2-2{\mathrm{x}}^{-3}\right)=2\times \left({\mathrm{x}}^2-{\mathrm{x}}^{-1}\right) \\ 2{\mathrm{x}}^2-2{\mathrm{x}}^{-1}=2{\mathrm{x}}^2-2{\mathrm{x}}^{-1} \end{gathered}$$

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

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 $\mathbf{(}\mathbf{0}\mathbf{,}\infty \mathbf{)}$.