solve the differential equation, subject to given condition

(x^2)(y')+1/y^2=0 , y(1)=2

any help would be appreciated. thanks!

i ended up with -x^-2=y^3 and im stuck but im pretty sure im doing it wrong =/

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- Mar 19th 2008, 12:44 PMs0urgrapeshelp with integration (differential equations)
solve the differential equation, subject to given condition

(x^2)(y')+1/y^2=0 , y(1)=2

any help would be appreciated. thanks!

i ended up with -x^-2=y^3 and im stuck but im pretty sure im doing it wrong =/ - Mar 19th 2008, 01:16 PMSoroban
Hello, s0urgrapes!

Quote:

Solve the differential equation, subject to given condition

. . $\displaystyle x^2\!\cdot\!\frac{dy}{dx} + \frac{1}{y^2} \:=\:0,\quad y(1)\,=\,2$

Separate variables: .$\displaystyle x^2\!\cdot\frac{dy}{dx} \:=\:-\frac{1}{y^2} \quad\Rightarrow\quad y^2dy \:=\:-x^{-2}dx$

Integrate: .$\displaystyle \frac{1}{3}y^3 \;=\;x^{-1} + C\quad\Rightarrow\quad y^3 \:=\:\frac{3}{x} + C$

From $\displaystyle y(1) = 2$, we have: .$\displaystyle 2^3 \:=\:\frac{3}{1} + C \quad\Rightarrow\quad C \,=\,5$

Therefore: . $\displaystyle \boxed{y^3 \;=\;\frac{3}{x} + 5}$

- Mar 19th 2008, 01:26 PMs0urgrapes
wow thanks alot

i think the mistake i made was making the -2 into a -3 while integrating instead of -1 like it should be