Practicing Success
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| What is the differential equation obtained by removing the arbitrary constants in the equation $y^2=a(b^2-x^2)$? |
$xyy''+xy'-yy'=0$ $xyy''+x(y')^2-yy'=0$ $xyy''+x(y')^2-y'=0$ $xy''+x(y')^2-yy'=0$ |
| $xyy''+x(y')^2-yy'=0$ |
| Differentiating w.r.to x we get $yy'=-ax$. Differentiating again w.r.to x we get $yy''+(y')^2=-a$. Dividing the second by first we get $\frac{(y')^2+yy''}{yy'}=\frac{-a}{-ax}$. Hence we get $xyy''+x(y')^2-yy'=0$ |
