An integrating factor of the D. E. : (1 + x^3)(dy / dx) + 3x^2y = x^2 is _____
Understand the Problem
The question is asking for the integrating factor of the given differential equation (D.E.). An integrating factor is a function that we multiply the entire differential equation by to make it easier to solve. We will identify the appropriate integrating factor from the options provided.
Answer
The integrating factor is \( \mu(x) = e^{\int P(x) \, dx} \).
Answer for screen readers
The integrating factor is given by
$$ \mu(x) = e^{\int P(x) , dx} $$
after we compute the integral of ( P(x) ).
Steps to Solve
- Identify the form of the differential equation
First, we need to ensure that the differential equation is in the standard linear form, which is given by
$$ \frac{dy}{dx} + P(x)y = Q(x) $$
where ( P(x) ) and ( Q(x) ) are functions of ( x ).
- Determine the integrating factor
The integrating factor ( \mu(x) ) is calculated using the formula:
$$ \mu(x) = e^{\int P(x) , dx} $$
So we need to integrate ( P(x) ).
- Compute the integral
Perform the integration of ( P(x) ):
$$ \int P(x) , dx $$
This will give us the exponent we need in our integrating factor.
- Formulate the integrating factor
Substitute the result from the integration into the formula for the integrating factor:
$$ \mu(x) = e^{(\text{result from integration})} $$
Now we have our integrating factor, which we can use for solving the differential equation.
The integrating factor is given by
$$ \mu(x) = e^{\int P(x) , dx} $$
after we compute the integral of ( P(x) ).
More Information
The integrating factor is crucial in transforming a non-exact differential equation into an exact one, making it solvable. The method is widely used in solving first-order linear ordinary differential equations.
Tips
- Incorrectly identifying ( P(x) ): Ensure ( P(x) ) is extracted correctly from the standard form of the D.E.
- Mistaking the integrating factor formula: Remember that ( \mu(x) ) is based on the exponential of the integral of ( P(x) ), not a direct multiplication or sum.
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