MHT CET · Maths · Differential Equations
The general solution of \(\sin ^{-1}\left(\frac{d y}{d x}\right)=x+y\) is
- A \(\tan (x+y)-\sec (x+y)=x^2+c\)
- B \(\tan (x+y)+\sec (x+y)=x^2+c\)
- C \(\tan (x+y)+\sec (x+y)=x+c\)
- D \(\tan (x+y)-\sec (x+y)=x+c\)
Answer & Solution
Correct Answer
(D) \(\tan (x+y)-\sec (x+y)=x+c\)
Step-by-step Solution
Detailed explanation
\(\sin ^{-1}\left(\frac{d y}{d x}\right)=x+y \)
\( \therefore \frac{d y}{d x}=\sin (x+y) \)
\( \text { Put } x+y=t \Rightarrow 1+\frac{d y}{d x}=\frac{d t}{d x} \)
\( \therefore \frac{d t}{d x}-1=\sin t \Rightarrow \frac{d t}{d x}=1+\sin t \)
\( \therefore \int \frac{d t}{1+\sin t}=\int d x \)
\( \therefore \int \frac{(1-\sin t) d t}{1-\sin } t=\int d x \Rightarrow \int \frac{1-\sin t}{\cos ^2 t} d t=\int d x \)
\( \Rightarrow \int\left(\sec ^2 t-\operatorname{sect} \tan t\right) d t=\int d x \)
\( \therefore \tan t-\sec t=x+c \)
\( \therefore \tan (x+y)-\sec (x+y)=x+c\)
\( \therefore \frac{d y}{d x}=\sin (x+y) \)
\( \text { Put } x+y=t \Rightarrow 1+\frac{d y}{d x}=\frac{d t}{d x} \)
\( \therefore \frac{d t}{d x}-1=\sin t \Rightarrow \frac{d t}{d x}=1+\sin t \)
\( \therefore \int \frac{d t}{1+\sin t}=\int d x \)
\( \therefore \int \frac{(1-\sin t) d t}{1-\sin } t=\int d x \Rightarrow \int \frac{1-\sin t}{\cos ^2 t} d t=\int d x \)
\( \Rightarrow \int\left(\sec ^2 t-\operatorname{sect} \tan t\right) d t=\int d x \)
\( \therefore \tan t-\sec t=x+c \)
\( \therefore \tan (x+y)-\sec (x+y)=x+c\)
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