`dy/dx + 2y/x = 3x-5` Solve the first-order differential equation

Monday, September 15, 2014

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}+{2}\frac{{y}}{{x}}={3}{x}-{5}$ Solve the first-order differential equation

Given $\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}+{2}\frac{{y}}{{x}}={3}{x}-{5}$

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

when the first order linear ordinary Differentian 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,

$\displaystyle{y}'+{2}\frac{{y}}{{x}}={3}{x}-{5}--------{\left({1}\right)}$

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

on comparing both we get,

$\displaystyle{p}{\left({x}\right)}=\frac{{2}}{{x}}{\quad\text{and}\quad}{q}{\left({x}\right)}={3}{x}-{5}$

so on solving with the above general solution we get:

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

= $\displaystyle\frac{{{\left(\int{{e}}^{{\int\frac{{2}}{{x}}{\left.{d}{x}\right.}}}\cdot{\left({3}{x}-{5}\right)}\right)}{\left.{d}{x}\right.}+{c}}}{{{e}}^{{\int\frac{{2}}{{x}}{\left.{d}{x}\right.}}}}$

first we shall solve

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

proceeding further, we get

y(x) = $\displaystyle\frac{{{\left(\int{{e}}^{{\int\frac{{2}}{{x}}{\left.{d}{x}\right.}}}\cdot{\left({3}{x}-{5}\right)}\right)}{\left.{d}{x}\right.}+{c}}}{{{e}}^{{\int\frac{{2}}{{x}}{\left.{d}{x}\right.}}}}$

= $\displaystyle\frac{{{\left(\int{{x}}^{{2}}\cdot{\left({3}{x}-{5}\right)}\right)}{\left.{d}{x}\right.}+{c}}}{{{x}}^{{2}}}$

= $\displaystyle\frac{{{\left(\int{\left({3}{{x}}^{{3}}-{5}{{x}}^{{2}}\right)}\right)}{\left.{d}{x}\right.}+{c}}}{{{x}}^{{2}}}$

= $\displaystyle\frac{{{3}\frac{{{x}}^{{4}}}{{4}}-{5}\frac{{{x}}^{{3}}}{{3}}+{c}}}{{{x}}^{{2}}}$

so $\displaystyle{y}{\left({x}\right)}=\frac{{{3}\frac{{{x}}^{{4}}}{{4}}-{5}\frac{{{x}}^{{3}}}{{3}}+{c}}}{{{{x}}^{{2}}}}$

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Monday, September 15, 2014

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}+{2}\frac{{y}}{{x}}={3}{x}-{5}$ Solve the first-order differential equation

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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?

Monday, September 15, 2014

Solve the first-order differential equation

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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?

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}+{2}\frac{{y}}{{x}}={3}{x}-{5}$ 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?