`ylnx - xy' = 0` Find the general solution of the differential equation

Thursday, January 7, 2016

$\displaystyle{y}{\ln{{x}}}-{x}{y}'={0}$ Find the general solution of the differential equation

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

to able to apply direct integration: $\displaystyle\int{f{{\left({y}\right)}}}{\left.{d}{y}\right.}=\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ .

Rearranging the problem:

$\displaystyle{y}{\ln{{\left({x}\right)}}}-{x}{y}'={0}$

$\displaystyle{y}{\ln{{\left({x}\right)}}}={x}{y}'$ or $\displaystyle{x}{y}'={y}{\ln{{\left({x}\right)}}}$

$\displaystyle\frac{{{x}{y}'}}{{{y}{x}}}=\frac{{{y}{\ln{{\left({x}\right)}}}}}{{{y}{x}}}$

$\displaystyle\frac{{{y}'}}{{y}}=\frac{{\ln{{\left({x}\right)}}}}{{x}}$

Applying direct integration, we denote $\displaystyle{y}'=\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}$ :

$\displaystyle\int\frac{{{y}'}}{{y}}=\int\frac{{\ln{{\left({x}\right)}}}}{{x}}$

$\displaystyle\int\frac{{1}}{{y}}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=\int\frac{{\ln{{\left({x}\right)}}}}{{x}}$

$\displaystyle\int\frac{{1}}{{y}}{\left({\left.{d}{y}\right.}\right)}=\int\frac{{\ln{{\left({x}\right)}}}}{{x}}{\left.{d}{x}\right.}$

For the left side, we apply the basic integration formula for logarithm: $\displaystyle\int\frac{{{d}{u}}}{{u}}={\ln}{\left|{u}\right|}+{C}$

$\displaystyle\int\frac{{1}}{{y}}{\left({\left.{d}{y}\right.}\right)}={\ln}{\left|{y}\right|}$

For the right side, we apply u-substitution by letting $\displaystyle{u}={\ln{{\left({x}\right)}}}$ then $\displaystyle{d}{u}=\frac{{1}}{{x}}{\left.{d}{x}\right.}$ .

$\displaystyle\int\frac{{\ln{{\left({x}\right)}}}}{{x}}{\left.{d}{x}\right.}=\int{u}{d}{u}$

Applying the Power Rule for integration : $\displaystyle\int{{x}}^{{n}}=\frac{{{x}}^{{{n}+{1}}}}{{{n}+{1}}}+{C}$ .

$\displaystyle\int{u}{d}{u}=\frac{{{u}}^{{{1}+{1}}}}{{{1}+{1}}}+{C}$

$\displaystyle=\frac{{{u}}^{{2}}}{{2}}+{C}$

Plug-in $\displaystyle{u}={\ln{{\left({x}\right)}}}$ in $\displaystyle\frac{{{u}}^{{2}}}{{2}}+{C}$ , we get:

$\displaystyle\int\frac{{\ln{{\left({x}\right)}}}}{{x}}{\left.{d}{x}\right.}=\frac{{{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}}{{2}}+{C}$

Combining the results, we get the general solution for differential equation $\displaystyle{\left({y}{\ln{{\left({x}\right)}}}-{x}{y}'={0}\right)}$ as:

$\displaystyle{\ln}{\left|{y}\right|}=\frac{{{\left({\ln}{\left|{x}\right|}\right)}}^{{2}}}{{2}}+{C}$

The general solution: $\displaystyle{\ln}{\left|{y}\right|}=\frac{{{\left({\ln}{\left|{x}\right|}\right)}}^{{2}}}{{2}}+{C}$ can be expressed as:

$\displaystyle{y}={C}_{{1}}{{e}}^{{\frac{{{\left({\ln}{\left|{x}\right|}\right)}}^{{2}}}{{2}}}}+{C}$ .

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Thursday, January 7, 2016

$\displaystyle{y}{\ln{{x}}}-{x}{y}'={0}$ 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?

Thursday, January 7, 2016

Find the general solution of the differential equation

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

$\displaystyle{y}{\ln{{x}}}-{x}{y}'={0}$ 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?