CUET · MATHS · PYQ PAPER 2023
Solution of the differential equation \(\log \left(\frac{d y}{d x}\right)=3 x+4 y\) given that y = 0 when x = 0 :
- A \(3 e^{3 x}+4 e^{-4 y}+7=0\)
- B \(-4 e^{3 x}+3 e^{-4 y}+7=0\)
- C \(4 e^{3 x}+3 e^{-4 y}-7=0\)
- D \(-3 e^{3 x}-4 e^{-4 y}+7=0\)
Answer & Solution
Correct Answer
(C) \(4 e^{3 x}+3 e^{-4 y}-7=0\)
Step-by-step Solution
Detailed explanation
\(\frac{dy}{dx} = e^{3x+4y} \implies e^{-4y} dy = e^{3x} dx\) \(\int e^{-4y} dy = \int e^{3x} dx \implies -\frac{1}{4} e^{-4y} = \frac{1}{3} e^{3x} + C\) \(y(0)=0 \implies -\frac{1}{4} e^0 = \frac{1}{3} e^0 + C \implies -\frac{1}{4} = \frac{1}{3} + C \implies C = -\frac{7}{12}\)…
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