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Math  /  Calculus

Questiondydx+Py=Q\frac{d y}{d x} + P y = Q Solve using integrating factors.

Studdy Solution

STEP 1

What is this asking? We're looking to solve a differential equation using a special trick called an integrating factor! Watch out! Don't forget to multiply every term by the integrating factor, not just the left side!

STEP 2

1. Identify the integrating factor
2. Multiply the equation by the integrating factor
3. Simplify and integrate
4. Solve for y y

STEP 3

Alright, let's get started!
We're dealing with a first-order linear differential equation of the form dydx+Py=Q\frac{d y}{d x} + P y = Q.
Our first mission is to **find the integrating factor**.
This magical factor is given by:
e∫P dxe^{\int P \, dx}Here, PP is the coefficient of yy in our equation.
So, let's integrate PP with respect to xx and then exponentiate it to get our integrating factor!

STEP 4

Once we've got our integrating factor e∫P dxe^{\int P \, dx}, the next step is to **multiply every term** in the original differential equation by this factor.
This is going to help us transform the left side into something we can easily integrate.
So, our equation becomes:
e∫P dx⋅dydx+e∫P dx⋅Py=e∫P dx⋅Qe^{\int P \, dx} \cdot \frac{d y}{d x} + e^{\int P \, dx} \cdot P y = e^{\int P \, dx} \cdot Q

STEP 5

Now, the left side of the equation should look like the derivative of a product.
Specifically, it should be the derivative of:
ddx(e∫P dx⋅y)\frac{d}{dx}\left( e^{\int P \, dx} \cdot y \right)This means we can rewrite the equation as:
ddx(e∫P dx⋅y)=e∫P dx⋅Q\frac{d}{dx}\left( e^{\int P \, dx} \cdot y \right) = e^{\int P \, dx} \cdot Q

STEP 6

The next step is to **integrate both sides** with respect to xx.
This will help us find the expression for yy.
So, we integrate:
∫ddx(e∫P dx⋅y) dx=∫e∫P dx⋅Q dx\int \frac{d}{dx}\left( e^{\int P \, dx} \cdot y \right) \, dx = \int e^{\int P \, dx} \cdot Q \, dx

STEP 7

After integrating, the left side simplifies to:
e∫P dx⋅ye^{\int P \, dx} \cdot yAnd the right side will be some function of xx, which we'll call F(x)F(x).
So, we have:
e∫P dx⋅y=F(x)+Ce^{\int P \, dx} \cdot y = F(x) + Cwhere CC is the constant of integration.

STEP 8

Finally, to **solve for yy**, divide both sides by the integrating factor:
y=F(x)+Ce∫P dxy = \frac{F(x) + C}{e^{\int P \, dx}}And there you have it!
We've found yy in terms of xx.

STEP 9

The solution to the differential equation dydx+Py=Q\frac{d y}{d x} + P y = Q using integrating factors is:
y=F(x)+Ce∫P dxy = \frac{F(x) + C}{e^{\int P \, dx}}where F(x)F(x) is the result of integrating e∫P dx⋅Qe^{\int P \, dx} \cdot Q with respect to xx.

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