The differential equation of all the ellipses | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The equation of an ellipse centered at the origin with axes as the coordinate axes is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. Differentiating

Vedclass · Accepted Answer

$x y y^{\prime \prime}+x y^{\prime 2}-y y^{\prime}=0$

Vedclass · Answer

(B) The equation of an ellipse centered at the origin with axes as the coordinate axes is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. Differentiating with respect to $x$, we get $\frac{2x}{a^{2}}+\frac{2y y^{\prime}}{b^{2}}=0$, which simplifies to $\frac{x}{a^{2}}+\frac{y y^{\prime}}{b^{2}}=0$. This gives $\frac{b^{2}}{a^{2}} = -\frac{y y^{\prime}}{x}$. Differentiating again with respect to $x$, we get $\frac{1}{a^{2}} + \frac{1}{b^{2}}(y y^{\prime \prime} + (y^{\prime})^{2}) = 0$. Substituting $\frac{1}{a^{2}} = -\frac{y y^{\prime}}{b^{2}x}$, we have $-\frac{y y^{\prime}}{b^{2}x} + \frac{1}{b^{2}}(y y^{\prime \prime} + (y^{\prime})^{2}) = 0$. Multiplying by $b^{2}x$, we get $-y y^{\prime} + x(y y^{\prime \prime} + (y^{\prime})^{2}) = 0$. Thus, $x y y^{\prime \prime} + x(y^{\prime})^{2} - y y^{\prime} = 0$.

The differential equation of all the ellipses centered at the origin and having axes as the coordinate axes is

Similar Questions

If $y = a + bx^2$,where $a$ and $b$ are arbitrary constants,then which of the following is true?

The differential equation of the family of lines having $x$-intercept as $a$ and $y$-intercept as $b$ is:

The differential equation corresponding to the family of ellipses $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$, where '$a$' is an arbitrary constant, is:

If $Ax^3+Bxy=4$ (where $A$ and $B$ are arbitrary constants) is the general solution of the differential equation $F(x) \frac{d^2 y}{d x^2}+G(x) \frac{d y}{d x}-2 y=0$,then $F(1)+G(1)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module