Match List-I with List-II :
|
List-I |
List-II |
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(A) Integrating factor of $x d y-\left(y+2 x^2\right) d x=0$ |
(I) $\frac{1}{x}$ |
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(B) Integrating factor of $\left(2 x^2-3 y\right) d x=x d y$ |
(II) $x$ |
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(C) Integrating factor of $\left(2 y+3 x^2\right) dx+x d y=0$ |
(III) $x^2$ |
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(D) Integrating factor of $2 x d y+\left(3 x^3+2 y\right) d x=0$ |
(IV) $x^3$ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (I), (B) - (IV), (C) - (III), (D) - (II)
(A) $x d y-\left(y+2 x^2\right) d x=0$
$\frac{dy}{dx}-\frac{y}{x}=2x$
$I.F.=e^{\int -\frac{1}{x}dx}=e^{-\log x}=\frac{1}{x}$ (I)
(B) $\left(2 x^2-3 y\right) d x=x d y$
$\frac{dy}{dx}+\frac{3y}{x}=2x$
so $I.F.=e^{\int\frac{3}{x}dx}=e^{3\log x}=x^3$ (IV)
(C) $\left(2 y+3 x^2\right) dx+x d y=0$
$\frac{dy}{dx}+\frac{2y}{x}=-3x$
so $I.F.=e^{\int\frac{2}{x}dx}=e^{2\log x}=x^2$ (III)
(D) $2 x d y+\left(3 x^3+2 y\right) d x=0$
$\frac{dy}{dx}+\frac{y}{x}=\frac{-3x^2}{2}$
$I.F.=e^{\int\frac{1}{x}dx}=e^{\log x}=x$ (II)