`(du)/(dv) = uvsin(v^2) , u(0) = 1` Find the particular solution that satisfies the initial condition

Wednesday, March 26, 2014

$\displaystyle\frac{{{d}{u}}}{{{d}{v}}}={u}{v}{\sin{{\left({{v}}^{{2}}\right)}}},{u}{\left({0}\right)}={1}$ Find the particular solution that satisfies the initial condition

An ordinary differential equation (ODE) has differential equation for a function with single variable. A first order ODE follows .

In the given problem: $\displaystyle\frac{{{d}{u}}}{{{d}{v}}}={u}{v}{\sin{{\left({{v}}^{{2}}\right)}}}$ , we may apply variable separable differential equation in a form of .

Divide both sides by "u" and cross-multiply dv to set it up as:

$\displaystyle\frac{{{d}{u}}}{{u}}={v}{\sin{{\left({{v}}^{{2}}\right)}}}{d}{v}.$

Apply direct integration: $\displaystyle\int\frac{{{d}{u}}}{{u}}=\int{v}{\sin{{\left({{v}}^{{2}}\right)}}}{d}{v}.$

For the left sign, we follow the basic integration formula for logarithm:

$\displaystyle\int\frac{{{d}{u}}}{{u}}={\ln}{\left|{u}\right|}$

For the right side, we follow the basic integration formula for sine function:

Let: $\displaystyle{w}={{v}}^{{2}}$ then $\displaystyle{d}{w}={2}{v}\cdot{d}{v}$ or $\displaystyle\frac{{{d}{w}}}{{2}}={v}{d}{v}$ .

The integral becomes:

$\displaystyle\int{v}{\sin{{\left({{v}}^{{2}}\right)}}}{d}{v}=\int{\sin{{\left({{v}}^{{2}}\right)}}}\cdot{v}{d}{v}$

$\displaystyle=\int{\sin{{\left({w}\right)}}}\cdot\frac{{{d}{w}}}{{2}}$

$\displaystyle={\left(\frac{{1}}{{2}}\right)}\int{\sin{{\left({w}\right)}}}{d}{w}$

$\displaystyle={\left(\frac{{1}}{{2}}\right)}\cdot{\left(-{\cos{{\left({w}\right)}}}\right)}+{C}$

$\displaystyle=-\frac{{\cos{{\left({w}\right)}}}}{{2}}+{C}$

Plug-in $\displaystyle{w}={{v}}^{{2}}$ on $\displaystyle-\frac{{\cos{{\left({w}\right)}}}}{{2}}+{C}$ , we get:

$\displaystyle\int{v}{\sin{{\left({{v}}^{{2}}\right)}}}{d}{v}=-\frac{{\cos{{\left({{v}}^{{2}}\right)}}}}{{2}}+{C}$

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

$\displaystyle{\ln}{\left|{u}\right|}=-\frac{{\cos{{\left({{v}}^{{2}}\right)}}}}{{2}}+{C}$

To solve for the arbitrary constant $\displaystyle{\left({C}\right)}$ , apply the initial condition $\displaystyle{u}{\left({0}\right)}={1}$ on $\displaystyle{\ln}{\left|{u}\right|}=-\frac{{\cos{{\left({{v}}^{{2}}\right)}}}}{{2}}+{C}$ :

$\displaystyle{\ln}{\left|{1}\right|}=-\frac{{\cos{{\left({{0}}^{{2}}\right)}}}}{{2}}+{C}$

$\displaystyle{0}=-\frac{{1}}{{2}}+{C}$

$\displaystyle{C}={0}+\frac{{1}}{{2}}$

$\displaystyle{C}=\frac{{1}}{{2}}$

Plug-in $\displaystyle{C}=\frac{{1}}{{2}}$ in $\displaystyle{\ln}{\left|{u}\right|}=-\frac{{\cos{{\left({{v}}^{{2}}\right)}}}}{{2}}+{C}$ , we get

$\displaystyle{\ln}{\left|{u}\right|}=-\frac{{\cos{{\left({{v}}^{{2}}\right)}}}}{{2}}+\frac{{1}}{{2}}$

$\displaystyle{u}={{e}}^{{-\frac{{\cos{{\left({{v}}^{{2}}\right)}}}}{{2}}+\frac{{1}}{{2}}}}$

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Wednesday, March 26, 2014

$\displaystyle\frac{{{d}{u}}}{{{d}{v}}}={u}{v}{\sin{{\left({{v}}^{{2}}\right)}}},{u}{\left({0}\right)}={1}$ 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?

Wednesday, March 26, 2014

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?

$\displaystyle\frac{{{d}{u}}}{{{d}{v}}}={u}{v}{\sin{{\left({{v}}^{{2}}\right)}}},{u}{\left({0}\right)}={1}$ 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?