Math Problem Statement

i need to do 8 , 10, and 11

i need to do 8, 10, and 11

Solution

Let's work through the differential equations you mentioned:

Problem 8: y=2y+x2+5y' = 2y + x^2 + 5 This is a linear first-order differential equation. We can solve it using the integrating factor method. We begin by writing it in the standard linear form: y2y=x2+5y' - 2y = x^2 + 5 The integrating factor, μ(x)\mu(x), is given by: μ(x)=e2dx=e2x\mu(x) = e^{\int -2 \, dx} = e^{-2x} Now, we multiply the entire equation by the integrating factor: e2xy2e2xy=e2x(x2+5)e^{-2x}y' - 2e^{-2x}y = e^{-2x}(x^2 + 5) The left-hand side is the derivative of e2xye^{-2x}y, so we have: ddx(e2xy)=e2x(x2+5)\frac{d}{dx} \left( e^{-2x} y \right) = e^{-2x}(x^2 + 5) Now integrate both sides: ddx(e2xy)dx=e2x(x2+5)dx\int \frac{d}{dx} \left( e^{-2x} y \right) \, dx = \int e^{-2x}(x^2 + 5) \, dx The integral on the right side will require integration by parts or other methods (we can break it down if you'd like). After solving, we'll get the general solution for y(x)y(x).


Problem 10: xdydx+2y=3x \frac{dy}{dx} + 2y = 3 Rearrange the equation to put it in standard linear form: dydx+2xy=3x\frac{dy}{dx} + \frac{2}{x}y = \frac{3}{x} The integrating factor is: μ(x)=e2xdx=e2lnx=x2\mu(x) = e^{\int \frac{2}{x} \, dx} = e^{2 \ln |x|} = x^2 Multiply the whole equation by the integrating factor: x2dydx+2xy=3xx^2 \frac{dy}{dx} + 2x y = 3x This simplifies to: ddx(x2y)=3x\frac{d}{dx}(x^2 y) = 3x Now, integrate both sides: ddx(x2y)dx=3xdx\int \frac{d}{dx}(x^2 y) \, dx = \int 3x \, dx The result of the integral will give you the general solution for y(x)y(x).


Problem 11: xdydx+4y=x3xx \frac{dy}{dx} + 4y = x^3 - x Rearrange the equation to the standard linear form: dydx+4xy=x21\frac{dy}{dx} + \frac{4}{x} y = x^2 - 1 The integrating factor is: μ(x)=e4xdx=e4lnx=x4\mu(x) = e^{\int \frac{4}{x} \, dx} = e^{4 \ln |x|} = x^4 Multiply through by the integrating factor: x4dydx+4x3y=x6x4x^4 \frac{dy}{dx} + 4x^3 y = x^6 - x^4 This simplifies to: ddx(x4y)=x6x4\frac{d}{dx}(x^4 y) = x^6 - x^4 Now integrate both sides: ddx(x4y)dx=(x6x4)dx\int \frac{d}{dx}(x^4 y) \, dx = \int (x^6 - x^4) \, dx After integration, we get the general solution for y(x)y(x).


Would you like me to go over the steps for these integrals in more detail, or should I just proceed to the final solutions?

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Math Problem Analysis

Mathematical Concepts

Differential Equations
First-Order Linear Differential Equations
Integrating Factor Method
Integration Techniques

Formulas

Integrating factor formula: μ(x) = e^∫P(x) dx
General solution formula: y(x) = (1/μ(x)) ∫(μ(x) * Q(x)) dx

Theorems

First-order linear differential equation solution theorem
Integration by parts

Suitable Grade Level

Undergraduate Mathematics