`12yy' - 7e^x = 0` Find the general solution of the differential equation

Tuesday, October 25, 2016

$\displaystyle{12}{y}{y}'-{7}{{e}}^{{x}}={0}$ Find the general solution of the differential equation

For the given problem: $\displaystyle{12}{y}{y}'-{7}{{e}}^{{x}}={0}$ , we can evaluate this by applying variable separable differential equation in which we express it in a form of $\displaystyle{f{{\left({y}\right)}}}{\left.{d}{y}\right.}={f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ .

Then, $\displaystyle{12}{y}{y}'-{7}{{e}}^{{x}}={0}$ can be rearrange into $\displaystyle{12}{y}{y}'={7}{{e}}^{{x}}$

Express $\displaystyle{y}'$ as $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}$ :

$\displaystyle{12}{y}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={7}{{e}}^{{x}}$

Apply direct integration in the form of $\displaystyle\int{f{{\left({y}\right)}}}{\left.{d}{y}\right.}=\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ :

$\displaystyle{12}{y}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={7}{{e}}^{{x}}$

$\displaystyle{12}{y}{\left.{d}{y}\right.}={7}{{e}}^{{x}}{\left.{d}{x}\right.}$

$\displaystyle\int{12}{y}{\left.{d}{y}\right.}=\int{7}{{e}}^{{x}}{\left.{d}{x}\right.}$

For the both side , we apply basic integration property: $\displaystyle\int{c}\cdot{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={c}\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$

$\displaystyle{12}\int{y}{\left.{d}{y}\right.}={7}\int{{e}}^{{x}}{\left.{d}{x}\right.}$

Applying Power Rule integration: $\displaystyle\int{{u}}^{{n}}{d}{u}=\frac{{{u}}^{{{n}+{1}}}}{{{n}+{1}}}$ on the left side.

$\displaystyle{12}\int{y}{\left.{d}{y}\right.}={12}\cdot\frac{{{y}}^{{{1}+{1}}}}{{{1}+{1}}}$

$\displaystyle=\frac{{{12}{{y}}^{{2}}}}{{2}}$

$\displaystyle={6}{{y}}^{{2}}$

Apply basic integration formula for exponential function: $\displaystyle\int{{e}}^{{u}}{d}{u}={{e}}^{{u}}+{C}$ on the right side.

$\displaystyle{7}\int{{e}}^{{x}}{\left.{d}{x}\right.}={7}{{e}}^{{x}}+{C}$

Combining the results for the general solution of differential equation:

$\displaystyle{6}{{y}}^{{2}}={7}{{e}}^{{x}}+{C}$

$\displaystyle\frac{{{6}{{y}}^{{2}}}}{{6}}=\frac{{{7}{{e}}^{{x}}}}{{6}}+{C}$

$\displaystyle{{y}}^{{2}}=\frac{{{7}{{e}}^{{x}}}}{{6}}+{C}$

$\displaystyle{y}=\pm\sqrt{{\frac{{{7}{{e}}^{{x}}}}{{6}}+{C}}}$

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Tuesday, October 25, 2016

$\displaystyle{12}{y}{y}'-{7}{{e}}^{{x}}={0}$ Find the general solution of the differential equation

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

Tuesday, October 25, 2016

Find the general solution of the differential equation

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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?

$\displaystyle{12}{y}{y}'-{7}{{e}}^{{x}}={0}$ Find the general solution of the differential equation

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