Math Problem Statement

Find the general solution of the differential equation.​ Then, use the initial condition to find the corresponding particular solution. xy'+3y=4x, y(1)=6

Solution

We are given the first-order linear differential equation:

xy+3y=4xxy' + 3y = 4x

and the initial condition y(1)=6y(1) = 6.

Step 1: Rewrite the equation in standard form

First, we divide the entire equation by xx to rewrite it in standard form:

y+3xy=4y' + \frac{3}{x} y = 4

This is a linear first-order differential equation of the form y+P(x)y=Q(x)y' + P(x)y = Q(x), where:

  • P(x)=3xP(x) = \frac{3}{x}
  • Q(x)=4Q(x) = 4

Step 2: Find the integrating factor

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

μ(x)=eP(x)dx=e3xdx=e3lnx=x3\mu(x) = e^{\int P(x) \, dx} = e^{\int \frac{3}{x} \, dx} = e^{3 \ln |x|} = |x|^3

Since x>0x > 0, we can simplify μ(x)=x3\mu(x) = x^3.

Step 3: Multiply the equation by the integrating factor

We now multiply the entire equation by the integrating factor x3x^3:

x3y+3x2y=4x3x^3 y' + 3x^2 y = 4x^3

The left-hand side is the derivative of x3yx^3 y:

ddx(x3y)=4x3\frac{d}{dx} (x^3 y) = 4x^3

Step 4: Integrate both sides

Integrating both sides with respect to xx:

x3y=4x3dx=x4+Cx^3 y = \int 4x^3 \, dx = x^4 + C

where CC is the constant of integration. Thus, the general solution is:

y=x4+Cx3=x+Cx3y = \frac{x^4 + C}{x^3} = x + \frac{C}{x^3}

Step 5: Apply the initial condition

We are given y(1)=6y(1) = 6. Substituting x=1x = 1 and y=6y = 6 into the general solution:

6=1+C136 = 1 + \frac{C}{1^3}

Solving for CC:

6=1+CC=56 = 1 + C \quad \Rightarrow \quad C = 5

Step 6: Write the particular solution

Substitute C=5C = 5 into the general solution to get the particular solution:

y=x+5x3y = x + \frac{5}{x^3}

Thus, the particular solution is:

y=x+5x3y = x + \frac{5}{x^3}


Would you like to explore any of the steps in more detail, or do you have any questions?

Here are 5 related questions for further practice:

  1. Solve xy+2y=3x2xy' + 2y = 3x^2 with y(1)=4y(1) = 4.
  2. Solve y2xy=x2y' - \frac{2}{x} y = x^2 with y(2)=1y(2) = 1.
  3. Find the general solution of y+y=x2y' + y = x^2.
  4. Solve the differential equation x2y+4xy=x5x^2 y' + 4xy = x^5.
  5. Solve y+5y=10xy' + 5y = 10x with the initial condition y(0)=2y(0) = 2.

Tip: For linear first-order differential equations, always look for an integrating factor to simplify the equation.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Linear First-Order Differential Equations
Integrating Factor

Formulas

Standard form of a linear first-order differential equation: y' + P(x)y = Q(x)
Integrating factor: μ(x) = e^∫P(x) dx
General solution: y = (x^4 + C)/x^3

Theorems

Method of Integrating Factors

Suitable Grade Level

Undergraduate - Differential Equations Course