JSM Learn - Q- 73. The general solution of the differential equation (ex + 1) ydy = (y + 1) ex dx  is: (A) (y + 1) = k (ex + 1)                      (B) y + 1 = ex + 1 + k (C) y = log {k (y + 1) (ex + 1)}          (D) <math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">y</mi><mo>=</mo><mi>log</mi><mfenced close="}" open="{"><mfrac><mfenced><mrow><msup><mi mathvariant="normal">e</mi><mi mathvariant="normal">x</mi></msup><mo>+</mo><mn>1</mn></mrow></mfenced><mrow><mi mathvariant="normal">y</mi><mo>+</mo><mn>1</mn></mrow></mfrac></mfenced><mo>+</mo><mi mathvariant="normal">k</mi></math>

Q- 73. The general solution of the differential equation (e^x + 1) ydy = (y + 1) e^x dx is:

(A) (y + 1) = k (e^x + 1) (B) y + 1 = e^x + 1 + k

Ans:(C)

Explanation:

$Given differential equation (e^{x} + 1) ydy = (y + 1) e^{x} dx \Rightarrow \frac{dy}{dx} = \frac{e^{x} [1 + y]}{[e^{x} + 1] y} \Rightarrow \frac{dx}{dy} = \frac{[e^{x} + 1] y}{e^{x} [1 + y]} \Rightarrow \frac{dx}{dy} = \frac{e^{x} y}{e^{x [1 + y]}} + \frac{y}{e^{x} [1 + y]} \Rightarrow \frac{dx}{dy} = \frac{y}{1 + y} + \frac{y}{[1 + y] e^{x}} \Rightarrow \frac{dx}{dy} = \frac{y}{1 + y} [1 + \frac{1}{e^{x}}] \Rightarrow \frac{dx}{dy} = \frac{y}{1 + y} [\frac{e^{x} + 1}{e^{x}}] \Rightarrow [\frac{y}{1 + y}] dy = [\frac{e^{x}}{e^{x} + 1}] dx \Rightarrow On integrating both sides, we get \Rightarrow \int \frac{y}{1 + y} dy = \int \frac{e^{x}}{e^{x} + 1} dx \Rightarrow \int \frac{1 + y - 1}{1 + y} dy = \int \frac{e^{x}}{e^{x} + 1} dx \Rightarrow \int dy - \int \frac{1}{1 + y} dy \Rightarrow = \int \frac{e^{x}}{e^{x} + 1} dx \Rightarrow y - \log |(1 + y)| = \log |(e^{x} + 1)| + \log k \Rightarrow y = \log (1 + y) + \log (1 + e^{x}) + \log k \Rightarrow y = \log \{k (1 + y) (1 + e^{x})\}$

Q- 73. The general solution of the differential equation (e^x + 1) ydy = (y + 1) e^x dx is:

(A) (y + 1) = k (e^x + 1) (B) y + 1 = e^x + 1 + k

(C) y = log {k (y + 1) (e^x + 1)} (D) $y = \log \{\frac{(e^{x} + 1)}{y + 1}\} + k$

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Q- 73. The general solution of the differential equation (ex + 1) ydy = (y + 1) ex dx is: (A) (y + 1) = k (ex + 1) (B) y + 1 = ex + 1 + k (C) y = log {k (y + 1) (ex + 1)} (D) y=logex+1y+1+k

Q- 73. The general solution of the differential equation (e^x + 1) ydy = (y + 1) e^x dx is:

(A) (y + 1) = k (e^x + 1) (B) y + 1 = e^x + 1 + k

(C) y = log {k (y + 1) (e^x + 1)} (D) $y = \log \{\frac{(e^{x} + 1)}{y + 1}\} + k$