If a x^2+2 h x y+b y^2=0,then frac{d^2 y}{d x | vedclass

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

Question

(D) Given equation: $a x^2+2 h x y+b y^2=0$. Differentiating with respect to $x$: $2 a x+2 h(y+x \frac{d y}{d x})+2 b y \frac{d y}{d x}=0$. Dividing b

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) Given equation: $a x^2+2 h x y+b y^2=0$. Differentiating with respect to $x$: $2 a x+2 h(y+x \frac{d y}{d x})+2 b y \frac{d y}{d x}=0$. Dividing by $2$: $a x+h y+h x \frac{d y}{d x}+b y \frac{d y}{d x}=0$. $\frac{d y}{d x}(h x+b y)=-(a x+h y) \implies \frac{d y}{d x}=-\frac{a x+h y}{h x+b y}$. Now,differentiate again with respect to $x$ using the quotient rule: $\frac{d^2 y}{d x^2} = -\frac{(h x+b y)(a+h \frac{d y}{d x})-(a x+h y)(h+b \frac{d y}{d x})}{(h x+b y)^2}$. Substituting $\frac{d y}{d x} = -\frac{a x+h y}{h x+b y}$: $\frac{d^2 y}{d x^2} = -\frac{(h x+b y)(a-h \frac{a x+h y}{h x+b y})-(a x+h y)(h-b \frac{a x+h y}{h x+b y})}{(h x+b y)^2}$. Simplifying the numerator: $\frac{d^2 y}{d x^2} = -\frac{(a h x+a b y-a h x-h^2 y)-(a h x+b h y-a b x-b h y)}{(h x+b y)^3} = -\frac{a b y-h^2 y-a h x+a b x}{(h x+b y)^3}$. Since $a x^2+2 h x y+b y^2=0$,we have $a x+h y = -\frac{h x+b y}{x} \cdot y$ (or use the property of homogeneous equations). The final result is $\frac{h^2-a b}{(h x+b y)^3} \cdot (-1)$ is not correct,the standard result for this specific homogeneous equation is $0$ because the equation represents two lines passing through the origin,so $y=mx$,thus $\frac{d^2 y}{d x^2}=0$.

If $a x^2+2 h x y+b y^2=0$,then $\frac{d^2 y}{d x^2}=$

Similar Questions

If $f(x) = b e^{ax} + a e^{bx}$,then $f^{\prime \prime}(0)$ is equal to

If $y = x + e^x$,then at $x = 1$,$\frac{d^2x}{dy^2}$ is equal to

If $y^2 = ax^2 + bx + c$,where $a, b, c$ are constants,then $y^3 \frac{d^2 y}{dx^2}$ is equal to

If $y = (\tan^{-1} 2x)^2 + (\cot^{-1} 2x)^2$,then $(1 + 4x^2)^2 y'' - 16 =$

Let $\cos ^{-1}\left(\frac{y}{b}\right)=\log _e\left(\frac{x}{n}\right)^n$. Then $A y_2+B y_1+C y=0$ is possible for:

Vedclass Test Series

Exam Paper Generator

Online Exam Module