Advanced Differential Equations Quiz

A differential equation dy/dx = f(x,y) is homogeneous if f(tx,ty) = f(x,y) for any t ≠ 0. For option C, replace x with tx and y with ty to get f(tx,ty) = ((tx)² + (ty)²)/((tx)(ty)) = (t² x² + t² y²)/(t² xy) = t² (x² + y²)/(t² xy) = (x² + y²)/(xy) = f(x,y). The other options do not satisfy this scaling property.

Since the equation is homogeneous of degree 1, the standard substitution for homogeneous equations is u = y/x, which means y = u x and dy/dx = u + x du/dx. This transforms the equation into a separable equation in terms of u and x.

Let u = y/x, so y = ux and dy/dx = u + x(du/dx). Substituting into the original equation gives us u + x(du/dx) = (x + 2ux)/(3x + ux) = x(1 + 2u)/x(3 + u) = (1 + 2u)/(3 + u). After isolating the derivative term, we get x(du/dx) = (1 + 2u)/(3 + u) - u = (1 + 2u)/(3 + u) - u(3 + u)/(3 + u). Then x(du/dx) = (1 + 2u - 3u - u²)/(3 + u) = (1 - u - u²)/(3 + u). Rearranging to separate variables finally gives (3 + u)/(1 - u - u²) du = dx/x

Let u = y/x, du/dx * x = sin u + u - u = sin u, so du/sin u = dx/x. We integrate both sides to obtain ln|tan(u/2)| = ln|x| + C. If we let C = ln|K| for some K, then we can write the equation as ln|tan(u/2)| = ln|x| + ln|K|. Exponentiating both sides gives tan(u/2) = Kx. Substituting back u = y/x gives the final solution tan(y/(2x)) = Kx (which is in the form of option C.

Rewrite as dy/dx = (y/x) + cos(y/x). Let v = y/x, then f(x,y) = y/x + cos(y/x) = v + cos v, and f(tx,ty) = (ty)/(tx) + cos(ty/tx) = v + cos v = f(x,y), so it is homogeneous.

dy/dx = (x² + y²)/(2 x y) = (x² + u² x²)/(2 x (u x)) = x² (1 + u²)/(2 u x²) = (1 + u²)/(2u). Then u + x du/dx = (1 + u²)/(2u), so x du/dx = (1 + u²)/(2u) - u.

The differential equation dy/dx = (y - 2x)/(y + x) is homogeneous. We use the substitution v = y/x, so that y = vx and dy/dx = v + x dv/dx. Substituting into the differential equation gives: v + x dv/dx = (v - 2)/(v + 1). Rearranging terms: x dv/dx = (v - 2)/(v + 1) - v = (v - 2 - v(v+1))/(v+1) = (v - 2 - v² - v)/(v+1) = (-v² - 2)/(v+1). Thus, we obtain the separable equation: (v+1)/(v²+2) dv = -dx/x. Now integrate both sides: ∫ (v+1)/(v²+2) dv = -∫ dx/x. The left integral splits into two parts: ∫ v/(v²+2) dv + ∫ 1/(v²+2) dv. The first integral is (1/2) ln(v²+2). The second integral is (1/√2) arctan(v/√2). Therefore, (1/2) ln(v²+2) + (1/√2) arctan(v/√2) = -ln|x| + C, where C is an arbitrary constant. Now substitute back v = y/x: (1/2) ln((y/x)² + 2) + (1/√2) arctan((y/x)/√2) = -ln|x| + C. Simplify the logarithmic term: (1/2) ln((y²+2x²)/x²) = (1/2) ln(y²+2x²) - ln|x|. So the equation becomes: (1/2) ln(y²+2x²) - ln|x| + (1/√2) arctan(y/(x√2)) = -ln|x| + C. After cancelling  the -ln|x| terms, we obtain the implicit solution: (1/2) ln(y²+2x²) + (1/√2) arctan(y/(x√2)) = C. Multiplying both sides by 2 gives an equivalent form: ln(y² + 2x²) + √2 arctan(y/(x√2)) = C.

Starting from u + x du/dx = (2+u)/(1+u), we isolate the term with du/dx by subtracting u from both sides: x du/dx = (2+u)/(1+u) - u. Combine the right side into a single fraction: write u as u(1+u)/(1+u) to get common denominator: (2+u - u(1+u))/(1+u) = (2+u - u - u²)/(1+u) = (2 - u²)/(1+u). So we have x du/dx = (2 - u²)/(1+u). Now separate variables by multiplying both sides by dx/x and also by (1+u)/(2-u²): (1+u)/(2-u²) du = dx/x.

For D, f(tx,ty) = (tx + 1)/(ty - 2) which is not equal to f(x,y) because of the constant terms +1 and -2 that do not scale with t.

For a homogeneous equation, dy/dx depends only on the ratio y/x. Since the ratio 6/4 is equal to 3/2, the value of f(x, y) is the same at both points. Therefore, the slopes are equal.

A differential equation is homogeneous if it can be written in the form dy/dx = f(y/x). The given equation simplifies to dy/dx = x + y/x. Because of the term 'x', the right-hand side is not a function of the ratio y/x alone. Alternatively, checking the scaling property, if f(x,y) = (x² + y)/x, then f(tx, ty) = ((tx)² + ty)/(tx) = (t²x² + ty)/(tx) = (tx² + y)/x, which is not equal to f(x,y). Therefore, the equation is not homogeneous.

Divide both sides by x dx (assuming x ≠ 0): dy/dx - (y/x) = x (1 + (y/x)²), which is homogeneous.

This is the standard method for solving such equations. If we let y = vx, the product rule for differentiation gives dy/dx = v + x(dv/dx). Substituting this into the original equation yields v + x(dv/dx) = f(v). This can be rearranged to x(dv/dx) = f(v) - v, which is separable as dv/(f(v) - v) = dx/x.

The presence of the transcendental term e^(x/y) makes the integration very difficult unless x/y is treated as a single variable v. If we used v = y/x, we would have e^(1/v), which is harder to handle in integration.

dy/dx = (x³ + y³)/(x y (x + y)) = (1 + (y/x)³)/((y/x) (1 + y/x)). Let u = y/x, then the right side becomes (1 + u³)/(u (1 + u)), which is a function of u only, confirming homogeneity and allowing separation after substitution.