Question:
For the differential equation
\[
\frac{d^3y}{dx^3}=0,
\]
\(y=ax^2+bx+c\) is
For the differential equation
\[
\frac{d^3y}{dx^3}=0,
\]
\(y=ax^2+bx+c\) is
Show Hint
For an \(n\)-th order differential equation, the general solution contains \(n\) arbitrary constants.
Updated On: Jun 26, 2026
- the general solution
- a particular solution
- not a solution
- a solution, but not a particular solution
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The Correct Option is A
Solution and Explanation
Step 1: Start with the given differential equation.
The given differential equation is \[ \frac{d^3y}{dx^3}=0. \] This means the third derivative of \(y\) with respect to \(x\) is zero.
Step 2: Integrate the differential equation.
Integrating once, \[ \frac{d^2y}{dx^2}=C_1. \] Integrating again, \[ \frac{dy}{dx}=C_1x+C_2. \] Integrating one more time, \[ y=\frac{C_1x^2}{2}+C_2x+C_3. \]
Step 3: Compare with the given expression.
The obtained solution can be written as \[ y=ax^2+bx+c, \] where \(a,b,c\) are arbitrary constants.
Since the third-order differential equation has three arbitrary constants, this represents the general solution.
Step 4: Final conclusion.
Therefore, \[ \boxed{\text{the general solution}} \]
The given differential equation is \[ \frac{d^3y}{dx^3}=0. \] This means the third derivative of \(y\) with respect to \(x\) is zero.
Step 2: Integrate the differential equation.
Integrating once, \[ \frac{d^2y}{dx^2}=C_1. \] Integrating again, \[ \frac{dy}{dx}=C_1x+C_2. \] Integrating one more time, \[ y=\frac{C_1x^2}{2}+C_2x+C_3. \]
Step 3: Compare with the given expression.
The obtained solution can be written as \[ y=ax^2+bx+c, \] where \(a,b,c\) are arbitrary constants.
Since the third-order differential equation has three arbitrary constants, this represents the general solution.
Step 4: Final conclusion.
Therefore, \[ \boxed{\text{the general solution}} \]
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