The differential equation formed by eliminati | vedclass

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

Question

(D) Given equation is $y=e^x(a \cos x+b \sin x)$.Differentiating with respect to $x$ using the product rule:$\frac{d y}{d x} = e^x(a \cos x + b \sin x

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Given equation is $y=e^x(a \cos x+b \sin x)$.Differentiating with respect to $x$ using the product rule:$\frac{d y}{d x} = e^x(a \cos x + b \sin x) + e^x(-a \sin x + b \cos x)$$\frac{d y}{d x} = y + e^x(-a \sin x + b \cos x) \quad \dots(I)$Differentiating again with respect to $x$:$\frac{d^2 y}{d x^2} = \frac{d y}{d x} + \frac{d}{d x}[e^x(-a \sin x + b \cos x)]$$\frac{d^2 y}{d x^2} = \frac{d y}{d x} + e^x(-a \sin x + b \cos x) + e^x(-a \cos x - b \sin x)$From equation $(I)$,$e^x(-a \sin x + b \cos x) = \frac{d y}{d x} - y$.Substituting this into the second derivative:$\frac{d^2 y}{d x^2} = \frac{d y}{d x} + (\frac{d y}{d x} - y) - e^x(a \cos x + b \sin x)$Since $e^x(a \cos x + b \sin x) = y$,we have:$\frac{d^2 y}{d x^2} = 2 \frac{d y}{d x} - y - y$$\frac{d^2 y}{d x^2} - 2 \frac{d y}{d x} + 2 y = 0$.

The differential equation formed by eliminating $a$ and $b$ from the equation $y=e^x(a \cos x+b \sin x)$ is

Similar Questions

If $\sqrt{y}=\cos ^{-1} x$,then it satisfies the differential equation $(1-x^{2}) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}=c$,where $c$ is equal to

The differential equation obtained by eliminating the arbitrary constant from the equation $y^2 = (x + c)^3$ is

The differential equation formed by eliminating $A$ and $B$ from $A x^2 + B y^2 = 1$ is

The differential equation of all circles passing through the origin and having their centers on the $x$-axis is

If $A$ and $B$ are arbitrary constants,then the differential equation having $y=A e^{-x}+B \cos x$ as its general solution is

Vedclass Test Series

Exam Paper Generator

Online Exam Module