`sqrt(x) + sqrt(y)y' = 0 , y(1) = 9` Find the particular solution that satisfies the initial condition

Friday, August 24, 2012

$\displaystyle\sqrt{{{x}}}+\sqrt{{{y}}}{y}'={0},{y}{\left({1}\right)}={9}$ Find the particular solution that satisfies the initial condition

For the given problem: $\displaystyle\sqrt{{{x}}}+\sqrt{{{y}}}{y}'={0},$ we may rearrange this to

$\displaystyle\sqrt{{{y}}}{y}'=-\sqrt{{{x}}}$

Recall that $\displaystyle{y}'$ is denoted as $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}$ then it becomes:

$\displaystyle\sqrt{{{y}}}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=-\sqrt{{{x}}}$

Apply the variable separable differential equation in a form of $\displaystyle{f{{\left({y}\right)}}}{\left.{d}{y}\right.}={g{{\left({x}\right)}}}{\left.{d}{x}\right.}$ .

$\displaystyle\sqrt{{{y}}}{\left({\left.{d}{y}\right.}\right)}=-\sqrt{{{x}}}{\left.{d}{x}\right.}$

Apply direct integration using the Power Rule: $\displaystyle\int{{u}}^{{n}}{d}{u}=\frac{{{u}}^{{{n}+{1}}}}{{{n}+{1}}}$ .

Note: $\displaystyle\sqrt{{{x}}}={{x}}^{{\frac{{1}}{{2}}}}{\quad\text{and}\quad}\sqrt{{{y}}}={{y}}^{{\frac{{1}}{{2}}}}$ .

$\displaystyle\int\sqrt{{{y}}}{\left({\left.{d}{y}\right.}\right)}=\int-\sqrt{{{x}}}{\left.{d}{x}\right.}$

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

$\displaystyle\frac{{{y}}^{{\frac{{1}}{{2}}+{1}}}}{{\frac{{1}}{{2}}+{1}}}=-\frac{{{x}}^{{\frac{{1}}{{2}}+{1}}}}{{\frac{{1}}{{2}}+{1}}}+{C}$

$\displaystyle\frac{{{y}}^{{\frac{{3}}{{2}}}}}{{\frac{{3}}{{2}}}}=-\frac{{{x}}^{{\frac{{3}}{{2}}}}}{{\frac{{3}}{{2}}}}+{C}$

$\displaystyle{{y}}^{{\frac{{3}}{{2}}}}\cdot\frac{{2}}{{3}}=-{{x}}^{{\frac{{3}}{{2}}}}\cdot\frac{{2}}{{3}}+{C}$

$\displaystyle\frac{{2}}{{3}}{{y}}^{{\frac{{3}}{{2}}}}=-\frac{{2}}{{3}}{{x}}^{{\frac{{3}}{{2}}}}+{C}$

The general solution of the differential equation is $\displaystyle\frac{{2}}{{3}}{{y}}^{{\frac{{3}}{{2}}}}=-\frac{{2}}{{3}}{{x}}^{{\frac{{3}}{{2}}}}+{C}$ .

Using the given initial condition $\displaystyle{y}{\left({1}\right)}={9}$ , we plug-in $\displaystyle{x}={1}$ and $\displaystyle{y}={9}$ to solve for C:

$\displaystyle\frac{{2}}{{3}}{{\left({9}\right)}}^{{\frac{{3}}{{2}}}}=-\frac{{2}}{{3}}\cdot{{1}}^{{\frac{{3}}{{2}}}}+{C}$

$\displaystyle\frac{{2}}{{3}}\cdot{27}=-\frac{{2}}{{3}}\cdot{1}+{C}$

$\displaystyle{18}=-\frac{{2}}{{3}}+{C}$

$\displaystyle{C}={18}+\frac{{2}}{{3}}$

$\displaystyle{C}=\frac{{56}}{{3}}$

So,

$\displaystyle\frac{{2}}{{3}}{{y}}^{{\frac{{3}}{{2}}}}=-\frac{{2}}{{3}}{{x}}^{{\frac{{3}}{{2}}}}+\frac{{56}}{{3}}$

$\displaystyle{{y}}^{{\frac{{3}}{{2}}}}={\left(\frac{{3}}{{2}}\right)}{\left(-\frac{{2}}{{3}}{{x}}^{{\frac{{3}}{{2}}}}+\frac{{56}}{{3}}\right)}$

$\displaystyle{{y}}^{{\frac{{3}}{{2}}}}=-{{x}}^{{\frac{{3}}{{2}}}}+{28}$

$\displaystyle{{\left({{y}}^{{\frac{{3}}{{2}}}}\right)}}^{{\frac{{2}}{{3}}}}={{\left(-{{x}}^{{\frac{{3}}{{2}}}}+{28}\right)}}^{{\frac{{2}}{{3}}}}$

$\displaystyle{y}={{\left(-{{x}}^{{\frac{{3}}{{2}}}}+{28}\right)}}^{{\frac{{2}}{{3}}}}$

y = $\displaystyle{\sqrt[{{3}}]{{{{\left(-{{x}}^{{\frac{{3}}{{2}}}}+{28}\right)}}^{{2}}}}}$

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Friday, August 24, 2012

$\displaystyle\sqrt{{{x}}}+\sqrt{{{y}}}{y}'={0},{y}{\left({1}\right)}={9}$ Find the particular solution that satisfies the initial condition

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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?

Friday, August 24, 2012

Find the particular solution that satisfies the initial condition

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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?

$\displaystyle\sqrt{{{x}}}+\sqrt{{{y}}}{y}'={0},{y}{\left({1}\right)}={9}$ 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?