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Question: Solve: $\frac{dy}{dx} = \frac{x+2y-3}{2x+y-3}$...

Solve: dydx=x+2y−32x+y−3\frac{dy}{dx} = \frac{x+2y-3}{2x+y-3}

Answer

x+y-2 = C(x-y)^3

Explanation

Solution

The given differential equation is of the form dydx=ax+by+ca′x+b′y+c′\frac{dy}{dx} = \frac{ax+by+c}{a'x+b'y+c'}.

  1. Check for Intersection of Lines: The lines are x+2y−3=0x+2y-3=0 and 2x+y−3=02x+y-3=0. Their slopes are −1/2-1/2 and −2-2, respectively. Since the slopes are different, the lines intersect.

  2. Find the Point of Intersection: Solving the system: h+2k−3=0h+2k-3=0 2h+k−3=02h+k-3=0 From the first equation, h=3−2kh = 3-2k. Substituting into the second: 2(3−2k)+k−3=0  ⟹  6−4k+k−3=0  ⟹  3−3k=0  ⟹  k=12(3-2k)+k-3=0 \implies 6-4k+k-3=0 \implies 3-3k=0 \implies k=1. Then h=3−2(1)=1h = 3-2(1) = 1. The point of intersection is (h,k)=(1,1)(h,k) = (1,1).

  3. Substitution: Let x=X+1x = X+1 and y=Y+1y = Y+1. Then dydx=dYdX\frac{dy}{dx} = \frac{dY}{dX}. Substituting into the differential equation: dYdX=(X+1)+2(Y+1)−32(X+1)+(Y+1)−3=X+1+2Y+2−32X+2+Y+1−3=X+2Y2X+Y\frac{dY}{dX} = \frac{(X+1)+2(Y+1)-3}{2(X+1)+(Y+1)-3} = \frac{X+1+2Y+2-3}{2X+2+Y+1-3} = \frac{X+2Y}{2X+Y}

  4. Solve the Homogeneous Equation: This is a homogeneous equation. Let Y=vXY = vX, so dYdX=v+XdvdX\frac{dY}{dX} = v + X\frac{dv}{dX}. v+XdvdX=X+2(vX)2X+(vX)=1+2v2+vv + X\frac{dv}{dX} = \frac{X+2(vX)}{2X+(vX)} = \frac{1+2v}{2+v} XdvdX=1+2v2+v−v=1+2v−v(2+v)2+v=1−v22+vX\frac{dv}{dX} = \frac{1+2v}{2+v} - v = \frac{1+2v-v(2+v)}{2+v} = \frac{1-v^2}{2+v}

  5. Separate Variables: 2+v1−v2dv=dXX\frac{2+v}{1-v^2} dv = \frac{dX}{X}

  6. Integrate Both Sides: Using partial fractions for the left side: 2+v(1−v)(1+v)=3/21−v+1/21+v\frac{2+v}{(1-v)(1+v)} = \frac{3/2}{1-v} + \frac{1/2}{1+v}. ∫(3/21−v+1/21+v)dv=∫1XdX\int \left(\frac{3/2}{1-v} + \frac{1/2}{1+v}\right) dv = \int \frac{1}{X} dX −32ln⁡∣1−v∣+12ln⁡∣1+v∣=ln⁡∣X∣+C1-\frac{3}{2}\ln|1-v| + \frac{1}{2}\ln|1+v| = \ln|X| + C_1 Multiplying by 2: −3ln⁡∣1−v∣+ln⁡∣1+v∣=2ln⁡∣X∣+2C1-3\ln|1-v| + \ln|1+v| = 2\ln|X| + 2C_1 ln⁡∣1+v(1−v)3∣=ln⁡(X2)+C2\ln\left|\frac{1+v}{(1-v)^3}\right| = \ln(X^2) + C_2 1+v(1−v)3=CX2\frac{1+v}{(1-v)^3} = C X^2

  7. Substitute Back: Substitute v=Y/Xv = Y/X: 1+Y/X(1−Y/X)3=CX2  ⟹  (X+Y)/X((X−Y)/X)3=CX2\frac{1+Y/X}{(1-Y/X)^3} = C X^2 \implies \frac{(X+Y)/X}{((X-Y)/X)^3} = C X^2 X+YX⋅X3(X−Y)3=CX2  ⟹  X2(X+Y)(X−Y)3=CX2\frac{X+Y}{X} \cdot \frac{X^3}{(X-Y)^3} = C X^2 \implies \frac{X^2(X+Y)}{(X-Y)^3} = C X^2 Assuming X≠0X \neq 0: X+Y(X−Y)3=C  ⟹  X+Y=C(X−Y)3\frac{X+Y}{(X-Y)^3} = C \implies X+Y = C(X-Y)^3 Substitute back X=x−1X = x-1 and Y=y−1Y = y-1: (x−1)+(y−1)=C((x−1)−(y−1))3(x-1) + (y-1) = C((x-1) - (y-1))^3 x+y−2=C(x−y)3x+y-2 = C(x-y)^3

    Note on Singular Solutions: The cases 1−v2=01-v^2=0 (i.e., v=1v=1 or v=−1v=-1) lead to singular solutions y=xy=x and y=−x+2y=-x+2. The solution y=−x+2y=-x+2 is included in the general solution when C=0C=0. The solution y=xy=x is a singular solution. The question asks for the general solution.