TS EAMCET · Maths · Differential Equations
The solution of the differential equation \(\frac{d y}{d x}=(4 x+y+1)^2\), when \(y(0)=1\) is
- A \(y=2 x^2-1-\frac{\pi}{8}\)
- B \(y=4 x-\left(1+\frac{\pi}{8}\right)\)
- C \(y=2 \tan \left(2 x+\frac{\pi}{4}\right)-4 x-1\)
- D \(y=2 \tan \left(x+\frac{\pi}{8}\right)+4 x-1\)
Answer & Solution
Correct Answer
(C) \(y=2 \tan \left(2 x+\frac{\pi}{4}\right)-4 x-1\)
Step-by-step Solution
Detailed explanation
We have, \(\frac{d y}{d x}=(4 x+y+1)^2\) Put \(4 x+y+1=v\) \(\begin{array}{rlrl}\Rightarrow & & 4+\frac{d y}{d x} & =\frac{d v}{d x} \\ \Rightarrow & \frac{d y}{d x} & =\frac{d v}{d x}-4\end{array}\) Now, \(\left(\frac{d v}{d x}-4\right)=v^2\)…
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