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 =/
Hello, s0urgrapes!
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: . $\displaystyle \boxed{y^3 \;=\;\frac{3}{x} + 5}$