MHT CET · Maths · Differential Equations
The particular solution of the differential equation \(\cos \left(\frac{\mathrm{dy}}{\mathrm{d} x}\right)=a\), under the conditions \(\mathrm{a} \epsilon \mathrm{R}\) and \(y(0)=2\) is
- A \(\cos \left(\frac{x-2}{y-2}\right)=a\)
- B \(\cos ^{-1}\left(\frac{y-2}{x}\right)=a\)
- C \(\cos \left(\frac{y-2}{x}\right)=a\)
- D \(\cos \left(\frac{x-2}{y+2}\right)=a\)
Answer & Solution
Correct Answer
(C) \(\cos \left(\frac{y-2}{x}\right)=a\)
Step-by-step Solution
Detailed explanation
\(\cos \left(\frac{d y}{d x}\right)=a \Rightarrow \frac{d y}{d x}=\cos ^{-1} a\)
\(\int d y=\int \cos ^{-1} a d x\)
\(\therefore y=x \cos ^{-1} a+c ....(1)\)
\(\text { We have } x=0, y=2\)
\(\therefore 2=0+c \Rightarrow c=2\)
\(\therefore y=x \cos ^{-1} a+2\)
\(\therefore y-2=x \cos ^{-1} a \Rightarrow \frac{y-2}{x}=\cos ^{-1}\) \(a \Rightarrow \cos \left(\frac{y-2}{x}\right)=a\)
\(\int d y=\int \cos ^{-1} a d x\)
\(\therefore y=x \cos ^{-1} a+c ....(1)\)
\(\text { We have } x=0, y=2\)
\(\therefore 2=0+c \Rightarrow c=2\)
\(\therefore y=x \cos ^{-1} a+2\)
\(\therefore y-2=x \cos ^{-1} a \Rightarrow \frac{y-2}{x}=\cos ^{-1}\) \(a \Rightarrow \cos \left(\frac{y-2}{x}\right)=a\)
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