# [SOLVED]1.5.7.9 find gereral solution

#### karush

##### Well-known member
$\tiny{1.5.7.9}$

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
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
$\tiny{1.5.7.9}$

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
$\tiny{1.5.7.9}$

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
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...
Mahalo