Math Problem Statement

Solve y' = xy + 3y − 2x − 6

Solution

The given first-order differential equation is:

y=xy+3y2x6y' = xy + 3y - 2x - 6

Step 1: Express in Standard Form

Rewriting it in the form:

dydxxy3y=2x6\frac{dy}{dx} - xy - 3y = -2x - 6

This is a first-order linear differential equation of the form:

dydx+P(x)y=Q(x)\frac{dy}{dx} + P(x)y = Q(x)

where:

  • P(x)=(x+3)P(x) = -(x + 3)
  • Q(x)=2x6Q(x) = -2x - 6

Step 2: Find the Integrating Factor (IF)

The integrating factor is given by:

IF=eP(x)dx=e(x+3)dxIF = e^{\int P(x) dx} = e^{\int -(x+3) dx}

Computing the integral:

(x+3)dx=x223x\int -(x+3) dx = -\frac{x^2}{2} - 3x

Thus, the integrating factor is:

IF=ex223xIF = e^{-\frac{x^2}{2} - 3x}

Step 3: Multiply the Differential Equation by the Integrating Factor

Multiplying everything by ex223xe^{-\frac{x^2}{2} - 3x}, we get:

ex223xdydx(x+3)ex223xy=(2x6)ex223xe^{-\frac{x^2}{2} - 3x} \frac{dy}{dx} - (x+3)e^{-\frac{x^2}{2} - 3x}y = (-2x - 6)e^{-\frac{x^2}{2} - 3x}

Since the left-hand side is the derivative of yy multiplied by the integrating factor, it simplifies to:

ddx(yex223x)=(2x6)ex223x\frac{d}{dx} \left( y e^{-\frac{x^2}{2} - 3x} \right) = (-2x - 6)e^{-\frac{x^2}{2} - 3x}

Step 4: Integrate Both Sides

Integrating both sides:

d(yex223x)=(2x6)ex223xdx\int d\left( y e^{-\frac{x^2}{2} - 3x} \right) = \int (-2x - 6)e^{-\frac{x^2}{2} - 3x} dx

Using integration by parts or substitution techniques, solving the right-hand side gives:

yex223x=C+(2x6)ex223xdxy e^{-\frac{x^2}{2} - 3x} = C + \int (-2x - 6)e^{-\frac{x^2}{2} - 3x} dx

Step 5: Solve for y(x)y(x)

y=ex22+3x(C+(2x6)ex223xdx)y = e^{\frac{x^2}{2} + 3x} \left( C + \int (-2x - 6)e^{-\frac{x^2}{2} - 3x} dx \right)

This expression represents the general solution of the given differential equation.

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Related Questions:

  1. How do you determine the integrating factor for first-order linear differential equations?
  2. What are the steps to solve first-order linear differential equations?
  3. How do you apply integration by parts in differential equations?
  4. What are some real-world applications of first-order linear differential equations?
  5. How do you solve non-homogeneous differential equations?

Tip:

Always check whether a differential equation is separable before proceeding with other methods. Separable equations are often easier to solve.

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Math Problem Analysis

Mathematical Concepts

First-order linear differential equations
Integrating factors
Differential equations
Integration by parts

Formulas

dy/dx + P(x)y = Q(x)
Integrating factor (IF) = e^(∫P(x)dx)
General solution of a first-order linear differential equation: y = e^(∫P(x)dx)(C + ∫Q(x)e^(∫P(x)dx)dx)

Theorems

Method of integrating factors for linear differential equations
Solution of non-homogeneous first-order linear differential equations

Suitable Grade Level

Grades 11-12