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Question Number 25981 by ifcrna380w last updated on 17/Dec/17

Answered by prakash jain last updated on 17/Dec/17

a^2 ((dy/dx))^2 =y  (dy/dx)=±((√y)/a)  (dy/dx)=((√y)/a)  ((ady)/(√y))=dx  2ay^(1/2) =x+c⇒y=(((x+c)^2 )/(4a^2 ))  similary for (dy/dx)=−((√y)/a)  y=(((x−c_1 )^2 )/(4a^2 ))

$${a}^{\mathrm{2}} \left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} ={y} \\ $$$$\frac{{dy}}{{dx}}=\pm\frac{\sqrt{{y}}}{{a}} \\ $$$$\frac{{dy}}{{dx}}=\frac{\sqrt{{y}}}{{a}} \\ $$$$\frac{{ady}}{\sqrt{{y}}}={dx} \\ $$$$\mathrm{2}{ay}^{\mathrm{1}/\mathrm{2}} ={x}+{c}\Rightarrow{y}=\frac{\left({x}+{c}\right)^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} } \\ $$$${similary}\:{for}\:\frac{{dy}}{{dx}}=−\frac{\sqrt{{y}}}{{a}} \\ $$$${y}=\frac{\left({x}−{c}_{\mathrm{1}} \right)^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} } \\ $$

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