Question:
The order of the differential equation whose general solution is \(y=a\sin x+b\cos x\), where \(a\) and \(b\) are arbitrary constants, is
The order of the differential equation whose general solution is \(y=a\sin x+b\cos x\), where \(a\) and \(b\) are arbitrary constants, is
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The number of arbitrary constants in the general solution gives the order of the differential equation.
Updated On: May 7, 2026
- \(2\)
- \(4\)
- \(1\)
- \(3\)
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The Correct Option is A
Solution and Explanation
We are given the general solution:
\[
y=a\sin x+b\cos x.
\]
Here, \(a\) and \(b\) are arbitrary constants.
In a differential equation, the number of arbitrary constants in the general solution is equal to the order of the differential equation.
Here there are two arbitrary constants:
\[
a \quad \text{and} \quad b.
\]
Therefore, the order of the differential equation is:
\[
2.
\]
We can also verify by differentiating.
Given:
\[
y=a\sin x+b\cos x.
\]
Differentiate once:
\[
y'=a\cos x-b\sin x.
\]
Differentiate again:
\[
y''=-a\sin x-b\cos x.
\]
But,
\[
a\sin x+b\cos x=y.
\]
So,
\[
y''=-y.
\]
Therefore,
\[
y''+y=0.
\]
This differential equation contains the second derivative \(y''\), so its order is \(2\).
Hence, the correct answer is:
\[
2.
\]
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