`dy/dx + y/x = 6x + 2` Solve the first-order differential equation

Sunday, November 30, 2014

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}+\frac{{y}}{{x}}={6}{x}+{2}$ Solve the first-order differential equation

Given $\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}+\frac{{y}}{{x}}={6}{x}+{2}$

when the first order linear ordinary differential equation has the form of

$\displaystyle{y}'+{p}{\left({x}\right)}{y}={q}{\left({x}\right)}$

then the general solution is ,

$\displaystyle{y}{\left({x}\right)}=\frac{{{\left(\int{{e}}^{{\int{p}{\left({x}\right)}{\left.{d}{x}\right.}}}\cdot{q}{\left({x}\right)}\right)}{\left.{d}{x}\right.}+{c}}}{{{e}}^{{\int{p}{\left({x}\right)}{\left.{d}{x}\right.}}}}$

so,

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}+\frac{{y}}{{x}}={6}{x}+{2}--------{\left({1}\right)}$

$\displaystyle{y}'+{p}{\left({x}\right)}{y}={q}{\left({x}\right)}---------{\left({2}\right)}$

on comparing both we get,

$\displaystyle{p}{\left({x}\right)}=\frac{{1}}{{x}}{\quad\text{and}\quad}{q}{\left({x}\right)}={6}{x}+{2}$

so on solving with the above general solution we get:

y(x)= $\displaystyle\frac{{{\left(\int{{e}}^{{\int{p}{\left({x}\right)}{\left.{d}{x}\right.}}}\cdot{q}{\left({x}\right)}\right)}{\left.{d}{x}\right.}+{c}}}{{{e}}^{{\int{p}{\left({x}\right)}{\left.{d}{x}\right.}}}}$

= $\displaystyle\frac{{{\left(\int{{e}}^{{\int{\left(\frac{{1}}{{x}}\right)}{\left.{d}{x}\right.}}}\cdot{\left({6}{x}+{2}\right)}\right)}{\left.{d}{x}\right.}+{c}}}{{{e}}^{{\int\frac{{1}}{{x}}{\left.{d}{x}\right.}}}}$

first we shall solve

$\displaystyle{{e}}^{{\int{\left(\frac{{1}}{{x}}\right)}{\left.{d}{x}\right.}}}={{e}}^{{\ln{{\left({x}\right)}}}}={\left|{x}\right|}$ as we know $\displaystyle\int{\left(\frac{{1}}{{x}}\right)}{\left.{d}{x}\right.}={\ln{{\left({x}\right)}}}$

so then we get $\displaystyle{{e}}^{{\int{\left(\frac{{1}}{{x}}\right)}{\left.{d}{x}\right.}}}={x}$

since x must be greater than 0, or else $\displaystyle{\ln{{\left({x}\right)}}}$ is undefined.

Proceeding further with

y(x) = $\displaystyle\frac{{{\left(\int{{e}}^{{\int{\left(\frac{{1}}{{x}}\right)}{\left.{d}{x}\right.}}}\cdot{\left({6}{x}+{2}\right)}\right)}{\left.{d}{x}\right.}+{c}}}{{{e}}^{{\int\frac{{1}}{{x}}{\left.{d}{x}\right.}}}}$

= $\displaystyle\frac{{\int{x}\cdot{\left({6}{x}+{2}\right)}{\left.{d}{x}\right.}+{c}}}{{x}}$

= $\displaystyle\frac{{\int{\left({6}{{x}}^{{2}}+{2}{x}\right)}{\left.{d}{x}\right.}+{c}}}{{x}}$

= $\displaystyle\frac{{{6}\frac{{{x}}^{{3}}}{{3}}+{2}\frac{{{x}}^{{2}}}{{2}}+{c}}}{{x}}$

= $\displaystyle\frac{{{2}{{x}}^{{3}}+{{x}}^{{2}}+{c}}}{{x}}$

So, $\displaystyle{y}=\frac{{{2}{{x}}^{{3}}+{{x}}^{{2}}+{c}}}{{x}}$

Blog

Sunday, November 30, 2014

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}+\frac{{y}}{{x}}={6}{x}+{2}$ Solve the first-order 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?

Sunday, November 30, 2014

Solve the first-order 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?

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}+\frac{{y}}{{x}}={6}{x}+{2}$ Solve the first-order differential equation

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