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Bernoulli鈥檚 equation

A first-order equation that can be written in the form $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{P}\left(\mathbf{x}\right)\mathbf{y}\mathbf{=}\mathbf{Q}\left(\mathbf{x}\right){\mathbf{y}}^{\mathbf{n}}$, where $\mathbf{P}\left(\mathbf{x}\right)$and $\mathbf{Q}\left(\mathbf{x}\right)$are continuous on an interval$\left(\mathbf{a}\mathbf{,}\mathbf{b}\right)$ and n is a real number, is called a Bernoulli equation.

Given $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\frac{\mathbf{y}}{\mathbf{x}\mathbf{-}\mathbf{2}}\mathbf{=}\mathbf{5}\left(\mathbf{x}\mathbf{-}\mathbf{2}\right){\mathbf{y}}^{\frac{1}{2}}$, .

Compare with the general form of the Bernoulli equation.

$\mathbf{n}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}},\mathbf{P}\left(\mathbf{x}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{x}\mathbf{-}\mathbf{2}}\mathrm{and}\mathbf{Q}\left(\mathbf{x}\right)\mathbf{=}\mathbf{5}\left(\mathbf{x}\mathbf{-}\mathbf{2}\right)$

Now, divide by ${\mathbf{y}}^{\frac{\mathbf{1}}{\mathbf{2}}}$, we get,

${\mathbf{y}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{x}\mathbf{-}\mathbf{2}}{\mathbf{y}}^{\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{=}\mathbf{5}\left(\mathbf{x}\mathbf{-}\mathbf{2}\right)路路路路路路路\left(1\right)$

Substitute $\mathbf{v}\mathbf{=}{\mathbf{y}}^{\frac{\mathbf{1}}{\mathbf{2}}}$.

Differentiate with respect to x.

$\begin{array}{c}\frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{y}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}}\frac{\mathbf{dy}}{\mathbf{dx}}\\ \mathbf{2}\frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{=}{\mathbf{y}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}}\frac{\mathbf{dy}}{\mathbf{dx}}\end{array}$

Substitute it on equation (1)

$\begin{array}{c}\mathbf{2}\frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{+}\frac{\mathbf{v}}{\mathbf{x}\mathbf{-}\mathbf{2}}\mathbf{=}\mathbf{5}\left(\mathbf{x}\mathbf{-}\mathbf{2}\right)\\ \frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{+}\frac{\mathbf{v}}{\mathbf{2}\left(\mathbf{x}\mathbf{-}\mathbf{2}\right)}\mathbf{=}\frac{\mathbf{5}\left(\mathbf{x}\mathbf{-}\mathbf{2}\right)}{\mathbf{2}}路路路路路路\left(2\right)\end{array}$

Now, integrate the first. Where .

$\mathbf{P}\left(\mathbf{x}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\left(\mathbf{x}\mathbf{-}\mathbf{2}\right)}$

$\begin{array}{c}鈭�\mathbf{P}\left(\mathbf{x}\right)\mathbf{dx}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}鈭�\frac{\mathbf{1}}{\mathbf{x}\mathbf{-}\mathbf{2}}\mathbf{dx}\\ \mathbf{=}\frac{\mathbf{ln}\left|\mathbf{x}\mathbf{-}\mathbf{2}\right|}{\mathbf{2}}\end{array}$

Then,

$\begin{array}{c}渭\left(\mathbf{x}\right)\mathbf{=}{\mathbf{e}}^{鈭�\mathbf{P}\left(\mathbf{x}\right)\mathbf{dx}}\\ \mathbf{=}{\mathbf{e}}^{\frac{\mathbf{ln}\left|\mathbf{x}\mathbf{-}\mathbf{2}\right|}{\mathbf{2}}}\\ \mathbf{=}\left|\sqrt{\mathbf{x}\mathbf{-}\mathbf{2}}\right|\end{array}$

Multiply$渭\left(\mathbf{x}\right)$ with equation (2).

$\begin{array}{c}\left|\sqrt{\mathbf{x}\mathbf{-}\mathbf{2}}\right|\frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{+}\frac{\mathbf{v}}{\mathbf{2}\left(\mathbf{x}\mathbf{-}\mathbf{2}\right)}\left|\sqrt{\mathbf{x}\mathbf{-}\mathbf{2}}\right|\mathbf{=}\frac{\mathbf{5}\left(\mathbf{x}\mathbf{-}\mathbf{2}\right)}{\mathbf{2}}\left|\sqrt{\mathbf{x}\mathbf{-}\mathbf{2}}\right|\\ \left|\sqrt{\mathbf{x}\mathbf{-}\mathbf{2}}\right|\frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}\sqrt{\mathbf{x}\mathbf{-}\mathbf{2}}}\mathbf{v}\mathbf{=}\frac{\mathbf{5}{\left(\mathbf{x}\mathbf{-}\mathbf{2}\right)}^{\frac{\mathbf{3}}{\mathbf{2}}}}{\mathbf{2}}\\ \frac{\mathbf{d}}{\mathbf{dx}}\left[\sqrt{\mathbf{x}\mathbf{-}\mathbf{2}}\mathbf{v}\right]\mathbf{=}\frac{\mathbf{5}{\left(\mathbf{x}\mathbf{-}\mathbf{2}\right)}^{\frac{\mathbf{3}}{\mathbf{2}}}}{\mathbf{2}}\end{array}$

Integrate both sides,

$\begin{array}{c}鈭�\frac{\mathbf{d}}{\mathbf{dx}}\left[\sqrt{\mathbf{x}\mathbf{-}\mathbf{2}}\mathbf{v}\right]\mathbf{dx}\mathbf{=}鈭�\frac{\mathbf{5}{\left(\mathbf{x}\mathbf{-}\mathbf{2}\right)}^{\frac{\mathbf{3}}{\mathbf{2}}}}{\mathbf{2}}\mathbf{dx}\\ \sqrt{\mathbf{x}\mathbf{-}\mathbf{2}}\mathbf{v}\mathbf{=}\frac{\mathbf{5}}{\mathbf{2}}鈭�{\left(\mathbf{x}\mathbf{-}\mathbf{2}\right)}^{\frac{\mathbf{3}}{\mathbf{2}}}\mathbf{dx}\end{array}$

Evaluate separately.

$鈭�{\left(\mathbf{x}\mathbf{-}\mathbf{2}\right)}^{\frac{\mathbf{3}}{\mathbf{2}}}\mathbf{dx}$

Let $\mathbf{u}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{2}鈫�\mathbf{dx}\mathbf{=}\mathbf{du}$

$\begin{array}{c}\sqrt{\mathbf{x}\mathbf{-}\mathbf{2}}\mathbf{v}\mathbf{=}\frac{\mathbf{5}}{\mathbf{2}}鈭�{\left(\mathbf{u}\right)}^{\frac{\mathbf{3}}{\mathbf{2}}}\mathbf{du}\\ \sqrt{\mathbf{x}\mathbf{-}\mathbf{2}}\mathbf{v}\mathbf{=}\frac{\mathbf{5}}{\mathbf{2}}\mathbf{脳}\frac{\mathbf{2}{\mathbf{u}}^{\frac{\mathbf{5}}{\mathbf{2}}}}{\mathbf{5}}\mathbf{+}{\mathbf{C}}_{\mathbf{1}}\\ \sqrt{\mathbf{x}\mathbf{-}\mathbf{2}}\mathbf{v}\mathbf{=}{\left(\mathbf{x}\mathbf{-}\mathbf{2}\right)}^{\frac{\mathbf{5}}{\mathbf{2}}}\mathbf{+}{\mathbf{C}}_{\mathbf{1}}\\ \mathbf{v}\mathbf{=}{\left(\mathbf{x}\mathbf{-}\mathbf{2}\right)}^{\mathbf{2}}\mathbf{+}\mathbf{C}{\left(\mathbf{x}\mathbf{-}\mathbf{2}\right)}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}}\end{array}$

Substitute $\mathbf{v}\mathbf{=}{\mathbf{y}}^{\frac{\mathbf{1}}{\mathbf{2}}}$.

$\begin{array}{c}{\mathbf{y}}^{\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{=}{\left(\mathbf{x}\mathbf{-}\mathbf{2}\right)}^{\mathbf{2}}\mathbf{+}\mathbf{C}{\left(\mathbf{x}\mathbf{-}\mathbf{2}\right)}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}}\\ \mathbf{y}\mathbf{=}{\left({\left(\mathbf{x}\mathbf{-}\mathbf{2}\right)}^{\mathbf{2}}\mathbf{+}\mathbf{C}{\left(\mathbf{x}\mathbf{-}\mathbf{2}\right)}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}}\right)}^{\mathbf{2}}\\ \mathbf{y}\mathbf{=}\frac{\mathbf{5}{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{x}}^{\mathbf{3}}\mathbf{+}\mathbf{C}}\end{array}$

Hence the solution is$\mathbf{y}\mathbf{=}\frac{\mathbf{5}{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{x}}^{\mathbf{3}}\mathbf{+}\mathbf{C}}$