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[SOLVED] 1.5.7.9 find gereral solution

karush

Well-known member
Jan 31, 2012
2,867
$\tiny{1.5.7.9}$

Find the general solution
$y'+\dfrac{3}{x^2}y=0$

$ u(x)=e^{\displaystyle\int \dfrac{3}{x^2} dx}=e^{-\dfrac{3}{x}}+c$

just seeing if this is started ok... not sure if the given eq should have been rewritten
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
699
You have misplaced the constant, c. It should also be in the exponential.

$\frac{dy}{dx}= -\frac{3}{x^2}y$
$\frac{dy}{y}= -\frac{3}{x^2}dx= -3x^{-2}dx$
Integrate both sides
$ln(y)= 3x^{-1}+ C= \frac{3}{x}+ C$
Take the exponential of both sides
$y= e^{\frac{3}{x}+ C}= e^Ce^{\frac{3}{x}}= C' e^{\frac{3}{x}}$
where $C'= e^C$.
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,429
$\tiny{1.5.7.9}$

Find the general solution
$y'+\dfrac{3}{x^2}y=0$

$ u(x)=e^{\displaystyle\int \dfrac{3}{x^2} dx}=e^{-\dfrac{3}{x}}+c$

just seeing if this is started ok... not sure if the given eq should have been rewritten
Since you are trying to use an integrating factor...

$\displaystyle \begin{align*} \mathrm{e}^{-\frac{3}{x}}\,y' + \frac{3}{x^2}\,\mathrm{e}^{-\frac{3}{x}}\,y &= 0 \\
\left( \mathrm{e}^{-\frac{3}{x}}\,y \right) ' &= 0 \\
\mathrm{e}^{-\frac{3}{x}}\,y &= \int{0\,\mathrm{d}x} \\
\mathrm{e}^{-\frac{3}{x}}\,y &= C \\
y &= C\,\mathrm{e}^{\frac{3}{x}} \end{align*} $
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,429
$\tiny{1.5.7.9}$

Find the general solution
$y'+\dfrac{3}{x^2}y=0$

$ u(x)=e^{\displaystyle\int \dfrac{3}{x^2} dx}=e^{-\dfrac{3}{x}}+c$

just seeing if this is started ok... not sure if the given eq should have been rewritten
Since you are trying to use an integrating factor...

$\displaystyle \begin{align*} \mathrm{e}^{-\frac{3}{x}}\,y' + \frac{3}{x^2}\,\mathrm{e}^{-\frac{3}{x}}\,y &= 0 \\
\left( \mathrm{e}^{-\frac{3}{x}}\,y \right) ' &= 0 \\
\mathrm{e}^{-\frac{3}{x}}\,y &= \int{0\,\mathrm{d}x} \\
\mathrm{e}^{-\frac{3}{x}}\,y &= C \\
y &= C\,\mathrm{e}^{\frac{3}{x}} \end{align*} $
 

karush

Well-known member
Jan 31, 2012
2,867
Since you are trying to use an integrating factor...

$\displaystyle \begin{align*} \mathrm{e}^{-\frac{3}{x}}\,y' + \frac{3}{x^2}\,\mathrm{e}^{-\frac{3}{x}}\,y &= 0 \\
\left( \mathrm{e}^{-\frac{3}{x}}\,y \right) ' &= 0 \\
\mathrm{e}^{-\frac{3}{x}}\,y &= \int{0\,\mathrm{d}x} \\
\mathrm{e}^{-\frac{3}{x}}\,y &= C \\
y &= C\,\mathrm{e}^{\frac{3}{x}} \end{align*} $
yes,, I came close to this but got the product wrong...:unsure:
Mahalo