`x^2 + 5y dy/dx = 0` Find the general solution of the differential equation

Thursday, May 17, 2012

$\displaystyle{{x}}^{{2}}+{5}{y}\frac{{\left.{d}{y}}}{{\left.{d}{x}}}={0}$ Find the general solution of the differential equation

Recall that an ordinary differential equation (ODE) has differential equation for a function with single variable. A first order ODE follows $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={f{{\left({x},{y}\right)}}}$ .

In the given problem: $\displaystyle{{x}}^{{2}}+{5}{y}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={0}$ , we apply variable separable differential equation in a form of $\displaystyle{f{{\left({y}\right)}}}{\left.{d}{y}\right.}={f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ .

Move the $\displaystyle{{x}}^{{2}}$ to the other side: $\displaystyle{5}{y}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=-{{x}}^{{2}}$

Transfer the $\displaystyle{\left({\left.{d}{x}\right.}\right)}$ to the other side by cross-multiplication: $\displaystyle{5}{y}{\left.{d}{y}\right.}=-{{x}}^{{2}}{\left.{d}{x}\right.}$

Apply direct integration: $\displaystyle\int{5}{y}{\left.{d}{y}\right.}=\int-{{x}}^{{2}}{\left.{d}{x}\right.}$

Apply the basic integration property: $\displaystyle\int{c}\cdot{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={c}\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ .

$\displaystyle{5}\int{y}{\left.{d}{y}\right.}={\left(-{1}\right)}\int{{x}}^{{2}}{\left.{d}{x}\right.}$

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

$\displaystyle{5}\cdot\frac{{{y}}^{{{1}+{1}}}}{{{1}+{1}}}={\left(-{1}\right)}\cdot\frac{{{x}}^{{{2}+{1}}}}{{{2}+{1}}}+{C}$

$\displaystyle\frac{{{5}{{y}}^{{2}}}}{{2}}=-\frac{{{x}}^{{3}}}{{3}}+{C}$

Multiply both side by 2/5, we get:

$\displaystyle{\left(\frac{{2}}{{5}}\right)}\frac{{{5}{{y}}^{{2}}}}{{2}}={\left(\frac{{2}}{{5}}\right)}{\left(-\frac{{{x}}^{{3}}}{{3}}+{C}\right)}$

Note: $\displaystyle{\left(\frac{{2}}{{5}}\right)}\cdot{C}={C}$ since $\displaystyle{C}$ is an arbitrary constant.

$\displaystyle{{y}}^{{2}}=\frac{{-{2}{{x}}^{{3}}}}{{15}}+{C}$

$\displaystyle{y}=\pm\sqrt{{-\frac{{{2}{{x}}^{{3}}}}{{15}}+{C}}}$

Blog

Thursday, May 17, 2012

$\displaystyle{{x}}^{{2}}+{5}{y}\frac{{\left.{d}{y}}}{{\left.{d}{x}}}={0}$ Find the general solution of the differential equation

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?

Thursday, May 17, 2012

Find the general solution of the differential equation

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?

$\displaystyle{{x}}^{{2}}+{5}{y}\frac{{\left.{d}{y}}}{{\left.{d}{x}}}={0}$ 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?