Math Problem Statement

dy/dx = y/x + x sin y/x

Solution

The given differential equation is:

dydx=yx+xsin(yx)\frac{dy}{dx} = \frac{y}{x} + x \sin\left(\frac{y}{x}\right)

Solution:

To solve this equation, let's attempt a substitution to simplify the equation. Define:

v=yxv = \frac{y}{x}

This implies:

y=vxanddydx=v+xdvdxy = vx \quad \text{and} \quad \frac{dy}{dx} = v + x\frac{dv}{dx}

Substituting these into the original equation:

v+xdvdx=v+xsin(v)v + x\frac{dv}{dx} = v + x \sin(v)

Canceling the vv terms from both sides:

xdvdx=xsin(v)x \frac{dv}{dx} = x \sin(v)

Dividing by xx on both sides:

dvdx=sin(v)\frac{dv}{dx} = \sin(v)

Now, separate variables:

dvsin(v)=dx\frac{dv}{\sin(v)} = dx

Integrating both sides:

dvsin(v)=dx\int \frac{dv}{\sin(v)} = \int dx

The integral of 1sin(v)\frac{1}{\sin(v)} is lntan(v/2)\ln|\tan(v/2)|, so we get:

lntan(v/2)=x+C\ln|\tan(v/2)| = x + C

Exponentiating both sides to remove the logarithm:

tan(v/2)=ex+C=AexwhereA=eC|\tan(v/2)| = e^{x+C} = A e^x \quad \text{where} \quad A = e^C

Thus:

tan(v/2)=±Aex\tan(v/2) = \pm A e^x

Finally, recall that v=yxv = \frac{y}{x}, so the solution is:

tan(y2x)=Aex\tan\left(\frac{y}{2x}\right) = A e^x

This is the implicit solution to the given differential equation.


Would you like further details or have any questions on this solution?

Related Questions:

  1. How can you verify if this solution satisfies the original equation?
  2. What is the significance of choosing the positive or negative sign in the solution?
  3. Can the solution be expressed explicitly in terms of yy?
  4. How does the constant AA affect the behavior of the solution?
  5. Can this method be applied to other types of differential equations?

Tip:

When facing a differential equation that involves yx\frac{y}{x}, substituting v=yxv = \frac{y}{x} is often a useful technique to simplify the problem.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Substitution Methods

Formulas

v = y/x
dy/dx = v + x dv/dx
dv/dx = sin(v)
Integral of 1/sin(v) = ln|tan(v/2)|

Theorems

Substitution in Differential Equations
Separation of Variables

Suitable Grade Level

Undergraduate Level (Calculus II or Differential Equations)