Math Problem Statement

Can you solve this? Please complete and proper the solution and make sure the answer is correc. And please guide me what i can write to my paper to avoid mistakes if im writing to paper. ( \cos x d x+\left(1+\frac{2}{y}\right) \sin x d y=0 )

Solution

Certainly! Let's solve the differential equation step by step:

Problem Statement

Solve the differential equation: cos(x)dx+(1+2y)sin(x)dy=0\cos(x) \, dx + \left(1 + \frac{2}{y}\right) \sin(x) \, dy = 0

Solution

  1. Separate Variables:
    We want to separate the variables xx and yy. Start by dividing both sides by sin(x)\sin(x) (assuming sin(x)0\sin(x) \neq 0): cos(x)sin(x)dx+(1+2y)dy=0\frac{\cos(x)}{\sin(x)} \, dx + \left(1 + \frac{2}{y}\right) \, dy = 0

    Simplify the equation: cot(x)dx+(1+2y)dy=0\cot(x) \, dx + \left(1 + \frac{2}{y}\right) \, dy = 0

  2. Rearrange Terms: To separate variables, move the terms involving yy to one side and the terms involving xx to the other: cot(x)dx=(1+2y)dy\cot(x) \, dx = -\left(1 + \frac{2}{y}\right) \, dy

  3. Integrate Both Sides:
    Now, integrate both sides of the equation separately.

    • For the left side: cot(x)dx=cos(x)sin(x)dx\int \cot(x) \, dx = \int \frac{\cos(x)}{\sin(x)} \, dx The integral of cot(x)\cot(x) is lnsin(x)\ln|\sin(x)|: lnsin(x)=C1\ln|\sin(x)| = \text{C}_1

    • For the right side: (1+2y)dy=1dy2ydy\int -\left(1 + \frac{2}{y}\right) \, dy = -\int 1 \, dy - \int \frac{2}{y} \, dy The integrals are: y2lny=C2-y - 2\ln|y| = \text{C}_2

  4. Combine the Results:
    The equation after integration becomes: lnsin(x)=y2lny+C\ln|\sin(x)| = -y - 2\ln|y| + C where C=C1+C2C = \text{C}_1 + \text{C}_2 is the constant of integration.

  5. Simplify the Expression:
    You can express the solution more compactly: lnsin(x)+y+2lny=C\ln|\sin(x)| + y + 2\ln|y| = C

    Alternatively, exponentiate both sides to eliminate the logarithms, if needed: sin(x)y2ey=eC|\sin(x)| \cdot y^2 \cdot e^y = e^C This can be simplified as: sin(x)y2ey=K|\sin(x)| \cdot y^2 \cdot e^y = K where K=eCK = e^C is another constant.

Final Answer

The general solution to the differential equation is: sin(x)y2ey=K|\sin(x)| \cdot y^2 \cdot e^y = K or equivalently: lnsin(x)+y+2lny=C\ln|\sin(x)| + y + 2\ln|y| = C where CC and KK are constants of integration.

Writing Tip

When writing this solution on paper:

  • Clearly separate each step.
  • Show the integration process explicitly, including the constants of integration.
  • Make sure to explain any assumptions, like sin(x)0\sin(x) \neq 0.
  • Write clearly and ensure all logarithmic and exponential steps are clearly stated.
  • Box or highlight the final answer to make it stand out.

Additional Questions to Consider:

  1. What if sin(x)=0\sin(x) = 0? How does it affect the solution?
  2. How would the solution change if the differential equation had different initial conditions?
  3. What are the implications of the constant CC in the general solution?
  4. Can this method be applied to similar differential equations with different functions?
  5. How would you solve the equation if cos(x)\cos(x) and sin(x)\sin(x) were replaced by other trigonometric functions?

Tip:

Always double-check each integration step to ensure constants and coefficients are correctly handled.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Variable Separation
Integration Techniques

Formulas

Integration by parts

Theorems

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Suitable Grade Level

Advanced Undergraduate