Math Problem Statement

Solve y' + y = 3x as a diff eq

Solution

The differential equation is:

y+y=3xy' + y = 3x

This is a first-order linear differential equation. We can solve it using an integrating factor. Here's the step-by-step process:

Step 1: Write the equation in standard form

The standard form of a first-order linear differential equation is:

y+P(x)y=Q(x)y' + P(x) y = Q(x)

In this case, we can identify:

  • P(x)=1P(x) = 1
  • Q(x)=3xQ(x) = 3x

Step 2: Find the integrating factor

The integrating factor μ(x)\mu(x) is given by:

μ(x)=eP(x)dx=e1dx=ex\mu(x) = e^{\int P(x) \, dx} = e^{\int 1 \, dx} = e^x

Step 3: Multiply the equation by the integrating factor

Multiplying both sides of the differential equation by exe^x:

exy+exy=3xexe^x y' + e^x y = 3x e^x

Step 4: Recognize the left-hand side as a derivative

The left-hand side of the equation can now be written as the derivative of the product yexy e^x:

ddx(yex)=3xex\frac{d}{dx} (y e^x) = 3x e^x

Step 5: Integrate both sides

Now, integrate both sides with respect to xx:

yex=3xexdxy e^x = \int 3x e^x \, dx

We can use integration by parts to solve the integral 3xexdx\int 3x e^x \, dx. Let:

  • u=xu = x, so du=dxdu = dx,
  • dv=exdxdv = e^x dx, so v=exv = e^x.

Now apply the integration by parts formula:

udv=uvvdu\int u \, dv = uv - \int v \, du

This gives:

3xexdx=3(xexexdx)=3(xexex)=3ex(x1)\int 3x e^x \, dx = 3 \left( x e^x - \int e^x \, dx \right) = 3(x e^x - e^x) = 3e^x(x - 1)

Thus, we have:

yex=3ex(x1)+Cy e^x = 3e^x(x - 1) + C

Step 6: Solve for yy

Divide both sides by exe^x:

y=3(x1)+Cexy = 3(x - 1) + C e^{-x}

Final solution:

The general solution to the differential equation is:

y(x)=3(x1)+Cexy(x) = 3(x - 1) + C e^{-x}

where CC is an arbitrary constant determined by initial conditions if provided.


Do you have any questions or would you like further details?

Related questions:

  1. What are integrating factors, and how do they work in solving differential equations?
  2. Can you show how to apply integration by parts more clearly?
  3. What are some examples of initial conditions that could specify CC here?
  4. How can this method be extended to non-linear differential equations?
  5. Are there any alternative methods to solve this type of equation?

Tip:

When solving first-order linear differential equations, always check if the equation is in standard form to apply the integrating factor method easily.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
First-order Linear Differential Equations
Integration by Parts
Integrating Factors

Formulas

Standard form of first-order linear differential equations: y' + P(x)y = Q(x)
Integrating factor: μ(x) = e^{∫P(x) dx}
Integration by parts: ∫u dv = uv - ∫v du

Theorems

First-order Linear Differential Equation Theorem
Integration by Parts

Suitable Grade Level

Undergraduate Calculus (Grade 12 or College-level)