Match List-I with List-II
|
List-I Differential Equations |
List-II Order and degree |
|
(A) $\frac{dy}{dx}+e^y = 0$ |
(I) order 2, degree not defined |
|
(B) $\frac{d^2y}{dx^2} =\left[1+(\frac{dy}{dx})^2\right]^{3/2}$ |
(II) order 2, degree 1 |
|
(C) $\left(\frac{d^2y}{dx^2}\right)^2+e^{(\frac{dy}{dx})}=0$ |
(III) order 1, degree 1 |
|
(D) $\frac{d^2y}{dx^2}+x\frac{dy}{dx}-2y= \log x;x>0$ |
(IV) order 2, degree 2 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (3) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
|
List-I Differential Equations |
List-II Order and degree |
|
(A) $\frac{dy}{dx}+e^y = 0$ |
(III) order 1, degree 1 |
|
(B) $\frac{d^2y}{dx^2} =\left[1+(\frac{dy}{dx})^2\right]^{3/2}$ |
(IV) order 2, degree 2 |
|
(C) $\left(\frac{d^2y}{dx^2}\right)^2+e^{(\frac{dy}{dx})}=0$ |
(I) order 2, degree not defined |
|
(D) $\frac{d^2y}{dx^2}+x\frac{dy}{dx}-2y= \log x;x>0$ |
(II) order 2, degree 1 |
$\text{(A)}\;\frac{dy}{dx}+e^{y}=0$
$\text{Order}=1,\;\text{Degree}=1$
$\text{(B)}\;\frac{d^{2}y}{dx^{2}}=\left[1+\left(\frac{dy}{dx}\right)^{2}\right]^{\frac{3}{2}}$
$\text{Order}=2,\;\text{Degree}=2$
$\text{(C)}\;\left(\frac{d^{2}y}{dx^{2}}\right)^{2}+e^{\frac{dy}{dx}}=0$
$\text{Order}=2,\;\text{Degree not defined}$
$\text{(D)}\;\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}-2y=\log x,\;x>0$
$\text{Order}=2,\;\text{Degree}=1$
$\text{Correct match: (A–III), (B–IV), (C–I), (D–II)}$