- #1

karush

Gold Member

MHB

- 3,269

- 5

$\tiny{b.2.2.1 \quad 48}$

320

solve $\quad y'=\dfrac{x^2}{y}$

$\begin{array}{lll}

\textit{Rewrite as}

&y\dfrac{dy}{dx}=x^2\implies y\ dy=x^2\ dx \\ \\

\textit{Integrate Thru}

&\displaystyle\int y \, dy = \displaystyle\int x^2 \, dx\\ \\

&\dfrac{y^2}{2} = \dfrac{x^3}{3}+c\\ \\

\textit{thus}

&3y^2-2x^3+c; \quad y\ne 0

\end{array}$

I think this is OK but always get confused about the c

320

solve $\quad y'=\dfrac{x^2}{y}$

$\begin{array}{lll}

\textit{Rewrite as}

&y\dfrac{dy}{dx}=x^2\implies y\ dy=x^2\ dx \\ \\

\textit{Integrate Thru}

&\displaystyle\int y \, dy = \displaystyle\int x^2 \, dx\\ \\

&\dfrac{y^2}{2} = \dfrac{x^3}{3}+c\\ \\

\textit{thus}

&3y^2-2x^3+c; \quad y\ne 0

\end{array}$

I think this is OK but always get confused about the c

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