Math Problem Statement

find the general solution of the differential equation y' = sin x + 2y

Solution

To find the general solution of the differential equation y=sinx+2yy' = \sin x + 2y, we will use the method of integrating factors. This is a first-order linear differential equation of the form:

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

where P(x)=2P(x) = -2 and Q(x)=sinxQ(x) = \sin x.

  1. Compute the integrating factor μ(x)\mu(x):

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

    μ(x)=eP(x)dx=e2dx=e2x\mu(x) = e^{\int P(x) \, dx} = e^{\int -2 \, dx} = e^{-2x}
  2. Multiply the entire differential equation by the integrating factor:

    e2xy+e2x2y=e2xsinxe^{-2x} y' + e^{-2x} \cdot 2y = e^{-2x} \sin x

    This simplifies to:

    (e2xy)=e2xsinx\left(e^{-2x} y\right)' = e^{-2x} \sin x
  3. Integrate both sides with respect to xx:

    e2xy=e2xsinxdxe^{-2x} y = \int e^{-2x} \sin x \, dx

    To integrate e2xsinxdx\int e^{-2x} \sin x \, dx, we use integration by parts. Let:

    u=sinxanddv=e2xdxu = \sin x \quad \text{and} \quad dv = e^{-2x} dx

    Then,

    du=cosxdxandv=e2xdx=12e2xdu = \cos x \, dx \quad \text{and} \quad v = \int e^{-2x} \, dx = -\frac{1}{2} e^{-2x}

    Now, apply integration by parts:

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

    Substituting uu and vv:

    e2xsinxdx=sinx(12e2x)(12e2x)cosxdx\int e^{-2x} \sin x \, dx = \sin x \left(-\frac{1}{2} e^{-2x}\right) - \int \left(-\frac{1}{2} e^{-2x}\right) \cos x \, dx

    Simplify:

    e2xsinxdx=12e2xsinx+12e2xcosxdx\int e^{-2x} \sin x \, dx = -\frac{1}{2} e^{-2x} \sin x + \frac{1}{2} \int e^{-2x} \cos x \, dx

    To solve e2xcosxdx\int e^{-2x} \cos x \, dx, we apply integration by parts again. Let:

    u=cosxanddv=e2xdxu = \cos x \quad \text{and} \quad dv = e^{-2x} dx

    Then,

    du=sinxdxandv=12e2xdu = -\sin x \, dx \quad \text{and} \quad v = -\frac{1}{2} e^{-2x}

    Apply integration by parts again:

    e2xcosxdx=cosx(12e2x)(12e2x)(sinx)dx\int e^{-2x} \cos x \, dx = \cos x \left(-\frac{1}{2} e^{-2x}\right) - \int \left(-\frac{1}{2} e^{-2x}\right) (-\sin x) \, dx

    Simplify:

    e2xcosxdx=12e2xcosx+12e2xsinxdx\int e^{-2x} \cos x \, dx = -\frac{1}{2} e^{-2x} \cos x + \frac{1}{2} \int e^{-2x} \sin x \, dx

    Combining the results of both parts, we get:

    I=e2xsinxdx=12e2xsinx12(12e2xcosx+12e2xsinxdx)I = \int e^{-2x} \sin x \, dx = -\frac{1}{2} e^{-2x} \sin x - \frac{1}{2} \left(-\frac{1}{2} e^{-2x} \cos x + \frac{1}{2} \int e^{-2x} \sin x \, dx\right)

    Solve for II:

    I=12e2xsinx+14e2xcosx14II = -\frac{1}{2} e^{-2x} \sin x + \frac{1}{4} e^{-2x} \cos x - \frac{1}{4} I 54I=12e2xsinx+14e2xcosx\frac{5}{4} I = -\frac{1}{2} e^{-2x} \sin x + \frac{1}{4} e^{-2x} \cos x I=25e2xsinx+15e2xcosxI = -\frac{2}{5} e^{-2x} \sin x + \frac{1}{5} e^{-2x} \cos x
  4. Include the constant of integration:

    e2xy=25e2xsinx+15e2xcosx+Ce^{-2x} y = -\frac{2}{5} e^{-2x} \sin x + \frac{1}{5} e^{-2x} \cos x + C
  5. Solve for yy:

    y=25sinx+15cosx+Ce2xy = -\frac{2}{5} \sin x + \frac{1}{5} \cos x + Ce^{2x}

Therefore, the general solution to the differential equation y=sinx+2yy' = \sin x + 2y is:

y=25sinx+15cosx+Ce2xy = -\frac{2}{5} \sin x + \frac{1}{5} \cos x + Ce^{2x}

Would you like any details or have any questions? Here are some related questions you might want to explore:

  1. How do you solve a differential equation using separation of variables?
  2. What are homogeneous and non-homogeneous differential equations?
  3. How do you find particular solutions to non-homogeneous differential equations?
  4. What are the applications of first-order differential equations in real life?
  5. How do you solve second-order linear differential equations with constant coefficients?

Tip: When solving differential equations, always ensure that you correctly identify the type of equation and use the appropriate method for the solution.

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Math Problem Analysis

Mathematical Concepts

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

Formulas

Integrating factor formula
Integration by parts formula

Theorems

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Suitable Grade Level

College