COMEDK · Maths · 32. Differential Equations
The particular solution of \(e^{\dfrac{d y}{d x}}=2 x+1\) given that \(y=1\) when \(x=0\) is
- A \(y=(x+1) \log |2 x+1|-x+1\)
- B \(y=\left(x+\dfrac{1}{2}\right) \log |2 x+1|-\dfrac{1}{2} x+1\)
- C \(y=\left(x-\dfrac{1}{2}\right) \log |2 x+1|-x-1\)
- D \(y=\left(x+\dfrac{1}{2}\right) \log |2 x+1|-x+1\)
Answer & Solution
Correct Answer
(D) \(y=\left(x+\dfrac{1}{2}\right) \log |2 x+1|-x+1\)
Step-by-step Solution
Detailed explanation
Given the differential equation \(e^{\dfrac{dy}{dx}} = 2x + 1\). Taking the natural logarithm on both sides, we get \(\dfrac{dy}{dx} = \ln(2x + 1)\). Integrating both sides with respect to \(x\), we have \(y = \int \ln(2x + 1) dx\). Let \(u = 2x + 1\), then \(du = 2 dx\), which…
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