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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 is a real number, is called a Bernoulli equation.

Given, $\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{+}{\mathbf{tx}}^{\mathbf{3}}\mathbf{+}\frac{\mathbf{x}}{\mathbf{t}}\mathbf{=}\mathbf{0}$.

Compare with the general form of the Bernoulli equation.

$\mathbf{n}\mathbf{=}3,\mathbf{P}\left(\mathbf{t}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{t}}\mathrm{and}\mathbf{Q}\left(\mathbf{t}\right)\mathbf{=}-\mathbf{t}$

Now, divide by ${\mathbf{x}}^3$, we get,

${\mathbf{x}}^{\mathbf{-}\mathbf{3}}\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{-}\mathbf{2}}}{\mathbf{t}}\mathbf{=}\mathbf{-}\mathbf{t}路路路路路路\left(1\right)$

Substitute $\mathbf{v}\mathbf{=}{\mathbf{x}}^{\mathbf{-}\mathbf{2}}$.

Differentiate with respect to t.

$\begin{array}{c}\frac{\mathbf{dv}}{\mathbf{dt}}\mathbf{=}\mathbf{-}\mathbf{2}{\mathbf{x}}^{\mathbf{-}\mathbf{3}}\frac{\mathbf{dx}}{\mathbf{dt}}\\ \mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\frac{\mathbf{dv}}{\mathbf{dt}}\mathbf{=}{\mathbf{x}}^{\mathbf{-}\mathbf{3}}\frac{\mathbf{dx}}{\mathbf{dt}}\end{array}$

Substitute it on equation (1)

$\begin{array}{c}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\frac{\mathbf{dv}}{\mathbf{dt}}\mathbf{+}\frac{\mathbf{v}}{\mathbf{t}}\mathbf{=}\mathbf{-}\mathbf{t}\\ \frac{\mathbf{dv}}{\mathbf{dt}}\mathbf{-}\frac{\mathbf{2}}{\mathbf{t}}\mathbf{v}\mathbf{=}\mathbf{2}\mathbf{t}路路路路路路\left(2\right)\end{array}$

Now, integrate $\mathbf{P}\left(\mathbf{t}\right)$the first. Where role="math" localid="1663935034942" $\mathbf{P}\left(\mathbf{x}\right)\mathbf{=}\mathbf{-}\frac{\mathbf{2}}{\mathbf{t}}$.

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

Then,

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

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

$\begin{array}{c}{\mathbf{t}}^{\mathbf{-}\mathbf{2}}\frac{\mathbf{dv}}{\mathbf{dt}}\mathbf{-}\mathbf{2}{\mathbf{t}}^{\mathbf{-}\mathbf{3}}\mathbf{v}\mathbf{=}\mathbf{2}{\mathbf{t}}^{\mathbf{-}\mathbf{1}}\\ \frac{\mathbf{d}}{\mathbf{dt}}\left[{\mathbf{t}}^{\mathbf{-}\mathbf{2}}\mathbf{v}\right]\mathbf{=}\mathbf{2}{\mathbf{t}}^{\mathbf{-}\mathbf{1}}\end{array}$

Integrate both sides,

$\begin{array}{c}鈭�\frac{\mathbf{d}}{\mathbf{dt}}\left[{\mathbf{t}}^{\mathbf{-}\mathbf{2}}\mathbf{v}\right]\mathbf{dt}\mathbf{=}鈭�\mathbf{2}{\mathbf{t}}^{\mathbf{-}\mathbf{1}}\mathbf{dt}\\ {\mathbf{t}}^{\mathbf{-}\mathbf{2}}\mathbf{v}\mathbf{=}\mathbf{2}鈭�\frac{\mathbf{1}}{\mathbf{t}}\mathbf{dt}\\ \mathbf{=}\mathbf{2}\mathbf{ln}\left|\mathbf{t}\right|\mathbf{+}{\mathbf{C}}_{\mathbf{1}}\\ \mathbf{v}\mathbf{=}\mathbf{2}{\mathbf{t}}^{\mathbf{2}}\mathbf{ln}\left|\mathbf{t}\right|\mathbf{+}{\mathbf{Ct}}^{\mathbf{2}}\end{array}$

Substitute$\mathbf{v}\mathbf{=}{\mathbf{x}}^{\mathbf{-}\mathbf{2}}$ .

${\mathbf{x}}^{\mathbf{-}\mathbf{2}}\mathbf{=}\mathbf{2}{\mathbf{t}}^{\mathbf{2}}\mathbf{ln}\left|\mathbf{t}\right|\mathbf{+}{\mathbf{Ct}}^{\mathbf{2}}$

Hence, the solution is${\mathbf{x}}^{\mathbf{-}\mathbf{2}}\mathbf{=}\mathbf{2}{\mathbf{t}}^{\mathbf{2}}\mathbf{ln}\left|\mathbf{t}\right|\mathbf{+}{\mathbf{Ct}}^{\mathbf{2}}$