Question
Find a solution to the differential equation dy/dx = 2x - 3y, and provide a step-by-step explanation of the method used to solve it.
Answers
GPT-4o mini
To solve the differential equation \(\frac{dy}{dx} = 2x - 3y\), we recognize that this is a first-order linear ordinary differential equation of the form:
\[
\frac{dy}{dx} + P(x)y = Q(x)
\]
where \(P(x) = 3\) and \(Q(x) = 2x\).
### Step 1: Identify the integrating factor
To solve this equation, we first need to determine an integrating factor, \(\mu(x)\). The integrating factor is given by:
\[
\mu(x) = e^{\int P(x) \, dx}
\]
Since \(P(x) = 3\):
\[
\mu(x) = e^{\int 3 \, dx} = e^{3x}
\]
### Step 2: Multiply through by the integrating factor
Next, we multiply the entire differential equation by the integrating factor \(e^{3x}\):
\[
e^{3x} \frac{dy}{dx} + e^{3x} \cdot 3y = e^{3x} \cdot 2x
\]
This can be rewritten as:
\[
\frac{d}{dx}(e^{3x}y) = 2xe^{3x}
\]
### Step 3: Integrate both sides
We now integrate both sides with respect to \(x\):
\[
\int \frac{d}{dx}(e^{3x}y) \, dx = \int 2xe^{3x} \, dx
\]
The left side simplifies to:
\[
e^{3x}y = \int 2xe^{3x} \, dx
\]
### Step 4: Solve the integral on the right side
To solve the integral \(\int 2xe^{3x} \, dx\), we can use integration by parts. Let:
- \(u = 2x\) \(\Rightarrow du = 2\,dx\)
- \(dv = e^{3x} dx\) \(\Rightarrow v = \frac{1}{3} e^{3x}\)
Now using integration by parts \(\int u \, dv = uv - \int v \, du\):
\[
\int 2x e^{3x} \, dx = 2x \cdot \frac{1}{3}e^{3x} - \int \frac{1}{3}e^{3x} \cdot 2\,dx
\]
\[
= \frac{2}{3}xe^{3x} - \frac{2}{3}\int e^{3x} \, dx
\]
\[
= \frac{2}{3}xe^{3x} - \frac{2}{3} \cdot \frac{1}{3}e^{3x} + C
\]
\[
= \frac{2}{3}xe^{3x} - \frac{2}{9}e^{3x} + C
\]
### Step 5: Substitute back
Substituting this result back into our equation gives:
\[
e^{3x}y = \frac{2}{3}xe^{3x} - \frac{2}{9}e^{3x} + C
\]
### Step 6: Solve for \(y\)
Now, divide every term by \(e^{3x}\):
\[
y = \frac{2}{3}x - \frac{2}{9} + Ce^{-3x}
\]
### Final solution
Thus, the general solution to the differential equation \(\frac{dy}{dx} = 2x - 3y\) is:
\[
y = \frac{2}{3}x - \frac{2}{9} + Ce^{-3x}
\]
where \(C\) is an arbitrary constant.
\[
\frac{dy}{dx} + P(x)y = Q(x)
\]
where \(P(x) = 3\) and \(Q(x) = 2x\).
### Step 1: Identify the integrating factor
To solve this equation, we first need to determine an integrating factor, \(\mu(x)\). The integrating factor is given by:
\[
\mu(x) = e^{\int P(x) \, dx}
\]
Since \(P(x) = 3\):
\[
\mu(x) = e^{\int 3 \, dx} = e^{3x}
\]
### Step 2: Multiply through by the integrating factor
Next, we multiply the entire differential equation by the integrating factor \(e^{3x}\):
\[
e^{3x} \frac{dy}{dx} + e^{3x} \cdot 3y = e^{3x} \cdot 2x
\]
This can be rewritten as:
\[
\frac{d}{dx}(e^{3x}y) = 2xe^{3x}
\]
### Step 3: Integrate both sides
We now integrate both sides with respect to \(x\):
\[
\int \frac{d}{dx}(e^{3x}y) \, dx = \int 2xe^{3x} \, dx
\]
The left side simplifies to:
\[
e^{3x}y = \int 2xe^{3x} \, dx
\]
### Step 4: Solve the integral on the right side
To solve the integral \(\int 2xe^{3x} \, dx\), we can use integration by parts. Let:
- \(u = 2x\) \(\Rightarrow du = 2\,dx\)
- \(dv = e^{3x} dx\) \(\Rightarrow v = \frac{1}{3} e^{3x}\)
Now using integration by parts \(\int u \, dv = uv - \int v \, du\):
\[
\int 2x e^{3x} \, dx = 2x \cdot \frac{1}{3}e^{3x} - \int \frac{1}{3}e^{3x} \cdot 2\,dx
\]
\[
= \frac{2}{3}xe^{3x} - \frac{2}{3}\int e^{3x} \, dx
\]
\[
= \frac{2}{3}xe^{3x} - \frac{2}{3} \cdot \frac{1}{3}e^{3x} + C
\]
\[
= \frac{2}{3}xe^{3x} - \frac{2}{9}e^{3x} + C
\]
### Step 5: Substitute back
Substituting this result back into our equation gives:
\[
e^{3x}y = \frac{2}{3}xe^{3x} - \frac{2}{9}e^{3x} + C
\]
### Step 6: Solve for \(y\)
Now, divide every term by \(e^{3x}\):
\[
y = \frac{2}{3}x - \frac{2}{9} + Ce^{-3x}
\]
### Final solution
Thus, the general solution to the differential equation \(\frac{dy}{dx} = 2x - 3y\) is:
\[
y = \frac{2}{3}x - \frac{2}{9} + Ce^{-3x}
\]
where \(C\) is an arbitrary constant.