MHT CET · Maths · Differential Equations
The solution of the differential equation \(\frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{y}{x}=\sin x\) is
- A \(x y+\cos x=\sin x+\mathrm{c}\), where \(\mathrm{c}\) is a constant of integration.
- B \(x(y+\cos x)=\sin x+\mathrm{c}\), where \(\mathrm{c}\) is a constant of integration.
- C \(y(x+\cos x)=\sin x+\mathrm{c}\), where \(\mathrm{c}\) is a constant of integration.
- D \(x y+\sin x=\cos x+c\), where \(\mathrm{c}\) is a constant of integration.
Answer & Solution
Correct Answer
(B) \(x(y+\cos x)=\sin x+\mathrm{c}\), where \(\mathrm{c}\) is a constant of integration.
Step-by-step Solution
Detailed explanation
For given linear differential equation,
\(\text { I.F. }=\mathrm{e}^{\int \frac{1}{x} d x}=\mathrm{e}^{\log x}=x\)
\(\therefore \quad\) The required solution is
\(y x=\int x \sin x \frac{\mathrm{d} y}{\mathrm{~d} x}\)
\(\begin{aligned} & \therefore \quad y x=-x \cos x+\int \cos x \mathrm{~d} x \\ & \therefore \quad y x=-x \cos x+\sin x+\mathrm{c} \\ & \therefore \quad x(y+\cos x)=\sin x+\mathrm{c}\end{aligned}\)
\(\text { I.F. }=\mathrm{e}^{\int \frac{1}{x} d x}=\mathrm{e}^{\log x}=x\)
\(\therefore \quad\) The required solution is
\(y x=\int x \sin x \frac{\mathrm{d} y}{\mathrm{~d} x}\)
\(\begin{aligned} & \therefore \quad y x=-x \cos x+\int \cos x \mathrm{~d} x \\ & \therefore \quad y x=-x \cos x+\sin x+\mathrm{c} \\ & \therefore \quad x(y+\cos x)=\sin x+\mathrm{c}\end{aligned}\)
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