Match List-I with List-II
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List-I Differential Equation |
List-II Integrating Factor |
|
(A) $\frac{dy}{dx}+2xy = 1$ |
(I) $x$ |
|
(B) $x\frac{dy}{dx}+2xy= 1$ |
(II) $e^{2x}$ |
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(C) $x\frac{dy}{dx}+y=1$ |
(III) $x^2$ |
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(D) $x\frac{dy}{dx}+2y=2$ |
(IV) $e^{x^2}$ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(IV), (B)-(II), (C)-(I), (D)-(III)
|
List-I Differential Equation |
List-II Integrating Factor |
|
(A) $\frac{dy}{dx}+2xy = 1$ |
(IV) $e^{x^2}$ |
|
(B) $x\frac{dy}{dx}+2xy= 1$ |
(II) $e^{2x}$ |
|
(C) $x\frac{dy}{dx}+y=1$ |
(I) $x$ |
|
(D) $x\frac{dy}{dx}+2y=2$ |
(III) $x^2$ |
$\text{For } \frac{dy}{dx}+P(x)\,y=Q(x),\ \text{I.F.}=e^{\int P(x)\,dx}$
$(A)\ \frac{dy}{dx}+2xy=1 \;\Rightarrow\; P(x)=2x \;\Rightarrow\; \text{I.F.}=e^{\int 2x\,dx}=e^{x^{2}}\ \Rightarrow\ (IV)$
$(B)\ x\frac{dy}{dx}+2xy=1 \;\Rightarrow\; \frac{dy}{dx}+2y=\frac{1}{x} \;\Rightarrow\; P(x)=2 \;\Rightarrow\; \text{I.F.}=e^{\int 2\,dx}=e^{2x}\ \Rightarrow\ (II)$
$(C)\ x\frac{dy}{dx}+y=1 \;\Rightarrow\; \frac{dy}{dx}+\frac{1}{x}y=\frac{1}{x} \;\Rightarrow\; P(x)=\frac{1}{x} \;\Rightarrow\; \text{I.F.}=e^{\int \frac{1}{x}\,dx}=x\ \Rightarrow\ (I)$
$(D)\ x\frac{dy}{dx}+2y=2 \;\Rightarrow\; \frac{dy}{dx}+\frac{2}{x}y=\frac{2}{x} \;\Rightarrow\; P(x)=\frac{2}{x} \;\Rightarrow\; \text{I.F.}=e^{\int \frac{2}{x}\,dx}=x^{2}\ \Rightarrow\ (III)$
Matching: $(A)\!\to\!(IV),\ (B)\!\to\!(II),\ (C)\!\to\!(I),\ (D)\!\to\!(III)$