The differential equation which represents th | vedclass

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

Question

(C) Given the family of curves: $y = c_1 e^{c_2 x} \dots (i)$Differentiating with respect to $x$,we get:$y' = c_1 c_2 e^{c_2 x} = c_2 y \dots (ii)$Aga

Vedclass · Accepted Answer

$y y'' = (y')^2$

Vedclass · Answer

(C) Given the family of curves: $y = c_1 e^{c_2 x} \dots (i)$Differentiating with respect to $x$,we get:$y' = c_1 c_2 e^{c_2 x} = c_2 y \dots (ii)$Again,differentiating with respect to $x$,we get:$y'' = c_2 y' \dots (iii)$From equation $(ii)$,we have $c_2 = \frac{y'}{y}$.Substituting this value of $c_2$ into equation $(iii)$:$y'' = \left( \frac{y'}{y} \right) y'$Multiplying both sides by $y$:$y y'' = (y')^2$This is the required differential equation.

The differential equation which represents the family of curves $y = c_1 e^{c_2 x}$,where $c_1$ and $c_2$ are arbitrary constants is:

Similar Questions

If the differential equation having $y=Ae^x+B \sin x$ as its general solution is $f(x) \frac{d^2 y}{d x^2}+g(x) \frac{d y}{d x}+h(x) y=0$,then $f(x)+g(x)+h(x)=$

$y=A e^x+B e^{-2 x}$ satisfies which of the following differential equations?

Verify that the given function $x^{2}=2 y^{2} \log y$ is a solution of the corresponding differential equation $(x^{2}+y^{2}) \frac{dy}{dx}-xy=0$.

The differential equation of all the lines in the $xy$-plane is

$y=A e^x+B e^{2 x}+C e^{3 x}$ satisfies the differential equation

Vedclass Test Series

Exam Paper Generator

Online Exam Module