TS EAMCET · Maths · Differential Equations
The solution of the differential equation \(\frac{d y}{d x}=\sin (x+y) \tan (x+y)-1\) is
- A \(\operatorname{cosec}(x+y)+\tan (x+y)=x+c\)
- B \(x+\operatorname{cosec}(x+y)=c\)
- C \(x+\tan (x+y)=c\)
- D \(x+\sec (x+y)=c\)
Answer & Solution
Correct Answer
(B) \(x+\operatorname{cosec}(x+y)=c\)
Step-by-step Solution
Detailed explanation
Given, \(\frac{d y}{d x}=\sin (x+y) \tan (x+y)-1\) \(\begin{aligned} & \text { Put } x+y=z \Rightarrow \quad 1+\frac{d y}{d x}=\frac{d z}{d x} \\ & \therefore \quad \frac{d z}{d x}-1=\sin z \tan z-1 \\ & \Rightarrow \quad \int \frac{\cos z}{\sin ^2 z} d z=\int d x\end{aligned}\)…
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