`dy/dx = 1/ sqrt(80+8x-16x^2)` Solve the differential equation

Monday, October 11, 2010

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{1}}{\sqrt{{{80}+{8}{x}-{16}{{x}}^{{2}}}}}$ Solve the differential equation

For the given differential equation: $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=\frac{{1}}{\sqrt{{{80}+{8}{x}-{16}{{x}}^{{2}}}}},$ we may write it in a form of $\displaystyle{N}{\left({y}\right)}{\left.{d}{y}\right.}={M}{\left({x}\right)}{\left.{d}{x}\right.}$ .

Cross-multiply the $\displaystyle{\left({\left.{d}{x}\right.}\right)}$ to the other side:

$\displaystyle{\left({\left.{d}{y}\right.}\right)}=\frac{{1}}{\sqrt{{{80}+{8}{x}-{16}{{x}}^{{2}}}}}{\left.{d}{x}\right.}$

To solve for the general solution of the differential equation, we may apply direct integration on both sides.

$\displaystyle\int{\left({\left.{d}{y}\right.}\right)}=\int\frac{{1}}{\sqrt{{{80}+{8}{x}-{16}{{x}}^{{2}}}}}{\left.{d}{x}\right.}$

For the left side, it follow basic integral formula:

$\displaystyle\int{\left.{d}{y}\right.}={y}$

To evaluate the right side, we may apply completing the square on the trinomial: $\displaystyle{80}+{8}{x}-{16}{{x}}^{{2}}=-{{\left({4}{x}-{1}\right)}}^{{2}}+{81}$ or $\displaystyle{81}-{{\left({4}{x}-{1}\right)}}^{{2}}$

Then, the integral on the right side becomes:

$\displaystyle\int\frac{{1}}{\sqrt{{{80}+{8}{x}-{16}{{x}}^{{2}}}}}{\left.{d}{x}\right.}=\int\frac{{1}}{\sqrt{{{81}-{{\left({4}{x}-{1}\right)}}^{{2}}}}}{\left.{d}{x}\right.}$

The integral resembles the basic integration formula for inverse sine function:

$\displaystyle\int\frac{{1}}{\sqrt{{{{a}}^{{2}}-{{u}}^{{2}}}}}{d}{u}={\arcsin{{\left(\frac{{u}}{{a}}\right)}}}+{C}$

We let $\displaystyle{u}={4}{x}-{1}$ then $\displaystyle{d}{u}={4}{\left.{d}{x}\right.}$ or $\displaystyle\frac{{{d}{u}}}{{4}}={\left.{d}{x}\right.}$ .

Note that $\displaystyle{81}={{9}}^{{2}}$

Then,

$\displaystyle\int\frac{{1}}{\sqrt{{{81}-{{\left({4}{x}-{1}\right)}}^{{2}}}}}{\left.{d}{x}\right.}=\int\frac{{1}}{\sqrt{{{{9}}^{{2}}-{{u}}^{{2}}}}}\cdot\frac{{{d}{u}}}{{4}}$

$\displaystyle={\left(\frac{{1}}{{4}}\right)}\int\frac{{1}}{\sqrt{{{{9}}^{{2}}-{{u}}^{{2}}}}}{d}{u}$

$\displaystyle={\left(\frac{{1}}{{4}}\right)}{\arcsin{{\left(\frac{{u}}{{9}}\right)}}}+{C}$

Plug-in $\displaystyle{u}={4}{x}-{1}$ in $\displaystyle{\left(\frac{{1}}{{4}}\right)}{\arcsin{{\left(\frac{{u}}{{9}}\right.}}}$ ), we get:

$\displaystyle\int\frac{{1}}{\sqrt{{{81}-{{\left({4}{x}-{1}\right)}}^{{2}}}}}{\left.{d}{x}\right.}={\left(\frac{{1}}{{4}}\right)}{\arcsin{{\left(\frac{{{4}{x}-{1}}}{{9}}\right)}}}+{C}$

Combining the results from both sides, we get the general solution of differential equation:

$\displaystyle{y}=\frac{{1}}{{4}}{\arcsin{{\left(\frac{{{4}{x}-{1}}}{{9}}\right)}}}+{C}$

Blog

Monday, October 11, 2010

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{1}}{\sqrt{{{80}+{8}{x}-{16}{{x}}^{{2}}}}}$ Solve the differential equation

No comments:

Post a Comment

Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

Monday, October 11, 2010

Solve the differential equation

No comments:

Post a Comment

Thomas Jefferson&#39;s election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{1}}{\sqrt{{{80}+{8}{x}-{16}{{x}}^{{2}}}}}$ Solve the differential equation

Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?