Question

Solve: \(\frac{dy}{dx} = \frac{x+y+1}{2x+2y+3}\).

Hint:

When the numerator and denominator of \(\frac{dy}{dx}\) are linear expressions of the form \(\frac{a_1x+b_1y+c_1}{a_2x+b_2y+c_2}\), check the ratio of coefficients \(\frac{a_1}{a_2}\) and \(\frac{b_1}{b_2}\). If they are equal, as in this case (\(\frac{1}{2}\)), a substitution like \(v=a_1x+b_1y\) will always transform the equation into a separable one.

Answer 1

Step 1: Understanding the Concept:
This is a first-order differential equation that is not directly separable, homogeneous, or linear. However, the terms involving x and y appear as a linear combination \(x+y\) in both the numerator and the denominator. This suggests that we can use a substitution to transform the equation into a separable form.
Step 2: Key Formula or Approach:
1. Let \(v = x+y\).
2. Differentiate this substitution with respect to x to find a replacement for \(\frac{dy}{dx}\).
3. Substitute both \(v\) and the expression for \(\frac{dy}{dx}\) into the original equation.
4. The resulting equation in terms of v and x will be separable. Solve it by integration.
5. Substitute back \(v = x+y\) to get the final solution.
Step 3: Detailed Explanation:
Let the substitution be: \[ v = x + y \] Differentiating with respect to x: \[ \frac{dv}{dx} = 1 + \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{dv}{dx} - 1 \] Substitute \(x+y=v\) and \(\frac{dy}{dx}\) into the given equation: \[ \frac{dv}{dx} - 1 = \frac{v+1}{2v+3} \] Now, solve for \(\frac{dv}{dx}\): \[ \frac{dv}{dx} = \frac{v+1}{2v+3} + 1 = \frac{(v+1) + (2v+3)}{2v+3} = \frac{3v+4}{2v+3} \] This is a separable equation. We can rearrange the terms to integrate: \[ \frac{2v+3}{3v+4} dv = dx \] Integrate both sides: \[ \int \frac{2v+3}{3v+4} dv = \int dx \] To solve the integral on the left, we can use algebraic manipulation: \[ \int \frac{2v+3}{3v+4} dv = \int \frac{\frac{2}{3}(3v+4) + 3 - \frac{8}{3}}{3v+4} dv = \int \left(\frac{2}{3} + \frac{1/3}{3v+4}\right) dv \] \[ = \frac{2}{3} \int dv + \frac{1}{3} \int \frac{1}{3v+4} dv \] \[ = \frac{2}{3}v + \frac{1}{3} \cdot \frac{\ln|3v+4|}{3} = \frac{2}{3}v + \frac{1}{9}\ln|3v+4| \] So, the equation becomes: \[ \frac{2}{3}v + \frac{1}{9}\ln|3v+4| = x + C_1 \] Substitute back \(v = x+y\): \[ \frac{2}{3}(x+y) + \frac{1}{9}\ln|3(x+y)+4| = x + C_1 \] Multiply by 9 to clear the fractions: \[ 6(x+y) + \ln|3x+3y+4| = 9x + 9C_1 \] \[ 6x + 6y + \ln|3x+3y+4| = 9x + C \] \[ 6y - 3x + \ln|3x+3y+4| = C \] Step 4: Final Answer:
The general solution of the differential equation is \(6y - 3x + \ln|3x + 3y + 4| = C\).