Question

Solve the following differential equations: (i) \(\frac{d y}{d x}=\frac{x+y+1}{x+y-1}\) (ii) \(\left(\frac{x+y-1}{x+y-2}\right) \frac{d y}{d x}=\frac{x+y+1}{x+y+2}\)

Short answer

Question: Solve the following first-order ODEs and find the general solution for y(x): (i) \(\frac{dy}{dx}=\frac{x+y+1}{x+y-1}\) (ii) \(\left(\frac{x+y-1}{x+y-2}\right) \frac{dy}{dx}=\frac{x+y+1}{x+y+2}\) Answer: (i) The general solution for the first ODE is: \(y = \frac{e^{2x+2C} - 2x - 1}{2}\) (ii) The general solution for the second ODE is: \(y = \frac{2x + 3 \pm (x+1)^{\frac{2}{3}}e^C}{2}\)

Step-by-step solution

Rearrange and simplify the given equation for better clarity

Rewrite the equation as: \(\frac{dy}{dx}=\frac{x+y+1}{x+y-1}\) Step 2: Substituting v = x+y

Introduce a variable substitution v = x+y

Define a new variable v such that v = x + y so that we can rewrite the equation as: \(\frac{dy}{dx}=\frac{v+1}{v-1}\) Step 3: Find dv/dx

Compute the differential of the new variable v with respect to x

Take the derivative of v with respect to x to find the relationship between dv and dx: \(\frac{dv}{dx}=\frac{d(x+y)}{dx}=1+\frac{dy}{dx}\) Step 4: Replace dy/dx in the equation

Use the relationship between dv/dx and dy/dx to rewrite the original differential equation

Substitute the expression for \(\frac{dy}{dx}\) in terms of \(\frac{dv}{dx}\): \(\frac{dv}{dx}-1=\frac{v+1}{v-1}\) Step 5: Separate variables

Separate dv terms on the left side and dx terms on the right side

Move the dx and dv terms to opposite sides of the equation to easily integrate both sides: \(\frac{dv}{(v+1)(v-1)}=dx\) Step 6: Integrate both sides

Integrate the separated variables equation with respect to v and x

Now, integrate both sides of the equation: \(\int\frac{dv}{(v+1)(v-1)}=\int dx\) Step 7: Apply partial fraction decomposition

Use partial fraction decomposition to rewrite the integral on the left-hand side

Expand the left-hand integral using partial fractions: \(\frac{A}{v-1}+\frac{B}{v+1}=\frac{1}{(v-1)(v+1)}\) Step 8: Solve for A and B

Equating coefficients to find A and B values

We have A + B = 0 and (A - B) = 1. Solving these, we get A = 1/2, B = -1/2. Step 9: Integrate the fractions

Integrate the right-hand side and the partial fractions on the left-hand side

Integrate each term: \(\int\frac{1/2}{v-1}dv-\int\frac{1/2}{v+1}dv=\int dx\) Which results in: \(\frac{1}{2}\ln\left|v-1\right|-\frac{1}{2}\ln\left|v+1\right|=x+C\) Step 10: Substitute back y and solve for y

Replace v with (x + y) and solve for y

Substitute v = x + y back into the equation: \(\frac{1}{2}\ln\left|(x+y)-1\right|-\frac{1}{2}\ln\left|(x+y)+1\right|=x+C\) Now, solve for y. (i) Solution: \(y = \frac{e^{2x+2C} - 2x - 1}{2}\) #(ii) Solving the second differential equation:# Step 1: Rewrite the equation

Rearrange and simplify the given equation for better clarity

Rewrite the equation as: \(\left(\frac{x+y-1}{x+y-2}\right) \frac{dy}{dx}=\frac{x+y+1}{x+y+2}\) Step 2: Cross-multiply both sides

Cross-multiply terms to simplify the equation

Multiply both sides of the equation to get: \((x+y+2)(x+y-2)\left[\frac{dy}{dx}\right]=(x+y-1)(x+y+1)\) Step 3: Simplify and write the equation in differential form

Simplify differential equation

Combine terms and simplify, then rewrite as: \(3\left[\frac{dy}{dx}\right]=(2x+3-2y)dy\) Step 4: Separate variables

Separate y terms on the left side and x terms on the right side of the differential

Move the y and x terms to opposite sides of the equation: \(\frac{dy}{(2x+3-2y)}=\frac{1}{3}\frac{dx}{x+1}\) Step 5: Integrate both sides

Integrate the separated variables equation with respect to y and x

Now, integrate both sides of the equation: \(\int\frac{dy}{(2x+3-2y)}=\frac{1}{3}\int\frac{dx}{x+1}\) Step 6: Simplify Integrals

Simplify constants and integrate both sides

Extract constants out of the integrals, giving: \(\int\frac{dy}{(2x+3-2y)}=\frac{1}{3}\int\frac{dx}{x+1}\) Step 7: Integrate and substitute variables

Integrate and substitute variables

Performing the integration, we obtain: \(-\frac{1}{2}\ln\left|2x+3-2y\right|=\frac{1}{3}\ln\left|x+1\right|+C\) Step 8: Solve for y

Solve for y

Now, exponentiate both sides and solve for y: \(\left|2x+3-2y\right|=\pm e^{C}\left(x+1\right)^{2/3}\) (ii) Solution: \(y = \frac{2x + 3 \pm (x+1)^{\frac{2}{3}}e^C}{2}\)