If l and m are respectively the order and the | vedclass

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

Question

(B) Given the general solution $y = a x^2 + b \log x$. First, differentiate with respect to $x$: $y' = 2ax + \frac{b}{x}$. Multiply by $x$: $x y' = 2a

Vedclass · Accepted Answer

$l$

Vedclass · Answer

(B) Given the general solution $y = a x^2 + b \log x$. First, differentiate with respect to $x$: $y' = 2ax + \frac{b}{x}$. Multiply by $x$: $x y' = 2ax^2 + b$. Differentiate again with respect to $x$: $x y'' + y' = 4ax$. From $y' = 2ax + \frac{b}{x}$, we have $b = x y' - 2ax^2$. Substitute $a = \frac{x y' - b}{2x^2}$ into the equation, or more simply, eliminate constants $a$ and $b$. From $y = ax^2 + b \log x$, we have $y' = 2ax + \frac{b}{x}$ and $y'' = 2a - \frac{b}{x^2}$. Solving for $a$ and $b$: $x^2 y'' = 2ax^2 - b$ and $y = ax^2 + b \log x$. Adding these: $x^2 y'' + y = 3ax^2 + b(\log x - 1)$. Actually, the differential equation is $x^2 y'' + x y' - 4y = 0$. Comparing with $f(x) y'' + g(x) y' = \frac{4y}{x}$, we get $f(x) = x^2$ and $g(x) = x$. The order $l = 2$ and the degree $m = 1$. Thus, $f(m) + g(m) = f(1) + g(1) = 1^2 + 1 = 2$. Since $l = 2$, $f(m) + g(m) = l$.

If $l$ and $m$ are respectively the order and the degree of the differential equation $f(x) y^{\prime \prime}+g(x) y^{\prime}=\frac{4 y}{x}$ whose general solution is $y=a x^2+b \log x$, then $f(m)+g(m)=$

Similar Questions

The degree of the differential equation $\left[1+\left(\frac{dy}{dx}\right)^{2}\right]^{2}=\frac{d^{2}y}{dx^{2}}$ is

The number of arbitrary constants in the particular solution of a differential equation of fourth order is . . . . . .

The order and degree of the differential equation $\left[1+\left(\frac{dy}{dx}\right)^{5}\right]^{\frac{1}{3}}=\frac{d^{2}y}{dx^{2}}$ are respectively

The order and degree of the differential equation $\sqrt{2 + (\frac{d^2y}{dx^2})^3} = (\frac{d^3y}{dx^3})^{5/2}$ are respectively:

The order and degree of the differential equation $\sqrt{1 + \frac{1}{(\frac{dy}{dx})^2}} = (\frac{d^2y}{dx^2})^{\frac{3}{2}}$ are ...

Vedclass Test Series

Exam Paper Generator

Online Exam Module