`2xy' - ln(x^2) = 0 , y(1) = 2` Find the particular solution that satisfies the initial condition

Sunday, April 6, 2014

$\displaystyle{2}{x}{y}'-{\ln{{\left({{x}}^{{2}}\right)}}}={0},{y}{\left({1}\right)}={2}$ Find the particular solution that satisfies the initial condition

The problem: $\displaystyle{2}{x}{y}'-{\ln{{\left({{x}}^{{2}}\right)}}}={0}$ is as first order ordinary differential equation that we can evaluate by applying variable separable differential equation:

$\displaystyle{N}{\left({y}\right)}{y}'={M}{\left({x}\right)}$

$\displaystyle{N}{\left({y}\right)}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={M}{\left({x}\right)}$

$\displaystyle{N}{\left({y}\right)}{\left.{d}{y}\right.}={M}{\left({x}\right)}{\left.{d}{x}\right.}$

Apply direct integration: $\displaystyle\int{N}{\left({y}\right)}{\left.{d}{y}\right.}=\int{M}{\left({x}\right)}{\left.{d}{x}\right.}$ to solve for the

general solution of a differential equation.

Then, $\displaystyle{2}{x}{y}'-{\ln{{\left({{x}}^{{2}}\right)}}}={0}$ will be rearrange in to $\displaystyle{2}{x}{y}'={\ln{{\left({{x}}^{{2}}\right)}}}$

Let $\displaystyle{y}'=\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}$ , we get: $\displaystyle{2}{x}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={\ln{{\left({{x}}^{{2}}\right)}}}$

or $\displaystyle{2}{x}{\left({\left.{d}{y}\right.}\right)}={\ln{{\left({{x}}^{{2}}\right)}}}{\left({\left.{d}{x}\right.}\right)}$

Divide both sides by $\displaystyle{x}$ to express in a form of $\displaystyle{N}{\left({y}\right)}{\left.{d}{y}\right.}={M}{\left({x}\right)}{\left.{d}{x}\right.}$ :

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

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

Applying direct integration, we will have:

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

For the left side, recall $\displaystyle\int{\left.{d}{y}\right.}={y}$ then $\displaystyle\int{2}{\left.{d}{y}\right.}={2}{y}$

For the right side, we let $\displaystyle{u}={{x}}^{{2}}$ then $\displaystyle{d}{u}={2}{x}{\left.{d}{x}\right.}$ or $\displaystyle{\left.{d}{x}\right.}=\frac{{{d}{u}}}{{{2}{x}}}$ .

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

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

$\displaystyle=\int\frac{{{\ln{{\left({u}\right)}}}{d}{u}}}{{{2}{u}}}$

$\displaystyle=\frac{{1}}{{2}}\int\frac{{\ln{{\left({u}\right)}}}}{{u}}{d}{u}$

Let $\displaystyle{v}={\ln{{\left({u}\right)}}}$ then $\displaystyle{d}{v}=\frac{{1}}{{u}}{d}{u}$ ,we get:

$\displaystyle\frac{{1}}{{2}}\int\frac{{\ln{{\left({u}\right)}}}}{{u}}{d}{u}=\frac{{1}}{{2}}\int{v}\cdot{d}{v}$

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

$\displaystyle\frac{{1}}{{2}}\int{v}\cdot{d}{v}=\frac{{1}}{{2}}\frac{{{v}}^{{{1}+{1}}}}{{{1}+{1}}}+{C}$

$\displaystyle=\frac{{1}}{{2}}\cdot\frac{{{v}}^{{2}}}{{2}}+{C}$

$\displaystyle=\frac{{1}}{{4}}{{v}}^{{2}}+{C}$

Recall $\displaystyle{v}={\ln{{\left({u}\right)}}}$ and $\displaystyle{u}={{x}}^{{2}}$ then $\displaystyle{v}={\ln{{\left({{x}}^{{2}}\right)}}}$ .

The integral will be:

$\displaystyle\int\frac{{{\ln{{\left({{x}}^{{2}}\right)}}}}}{{x}}{\left.{d}{x}\right.}=\frac{{1}}{{4}}{{\left({\ln{{\left({{x}}^{{2}}\right)}}}\right)}}^{{2}}+{C}{\quad\text{or}\quad}\frac{{{\left({\ln{{\left({{x}}^{{2}}\right)}}}\right)}}^{{2}}}{{4}}+{C}$

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

$\displaystyle{2}{y}=\frac{{{\left({\ln{{\left({{x}}^{{2}}\right)}}}\right)}}^{{2}}}{{4}}+{C}$

or $\displaystyle{y}=\frac{{{\left({\ln{{\left({{x}}^{{2}}\right)}}}\right)}}^{{2}}}{{8}}+{C}$

To solve for the arbitary constant (C), we consider the initial condition $\displaystyle{y}{\left({1}\right)}={2}$

When we plug-in the values, we get:

$\displaystyle{2}=\frac{{{\left({\ln{{\left({{1}}^{{2}}\right)}}}\right)}}^{{2}}}{{8}}+{C}$

$\displaystyle{2}=\frac{{0}}{{8}}+{C}$

$\displaystyle{2}={0}+{C}$

then $\displaystyle{C}={2}$

.Plug-in $\displaystyle{C}={2}$ on the general solution: $\displaystyle{y}=\frac{{{\left({\ln{{\left({{x}}^{{2}}\right)}}}\right)}}^{{2}}}{{8}}+{C}$ , we get the

particular solution as:

$\displaystyle{y}=\frac{{{\left({\ln{{\left({{x}}^{{2}}\right)}}}\right)}}^{{2}}}{{8}}+{2}$

Blog

Sunday, April 6, 2014

$\displaystyle{2}{x}{y}'-{\ln{{\left({{x}}^{{2}}\right)}}}={0},{y}{\left({1}\right)}={2}$ Find the particular solution that satisfies the initial condition

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?

Sunday, April 6, 2014

Find the particular solution that satisfies the initial condition

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?

$\displaystyle{2}{x}{y}'-{\ln{{\left({{x}}^{{2}}\right)}}}={0},{y}{\left({1}\right)}={2}$ Find the particular solution that satisfies the initial condition

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