Solve the differential equation dy/dx = 3 - 4y.

Step-by-step solution

1. 1

   Recognise the type and spot the equilibrium. The right side depends on $y$ only, so the equation is separable (it is also first-order linear). Setting $3-4y=0$ gives the constant solution $y=\tfrac34$ — the level at which the derivative vanishes. Every other solution will move toward or away from it, so this value should appear in the final answer.

2. 2

   Separate the variables. Divide by $3-4y$ (valid while $y\neq\tfrac34$) and multiply by $dx$:

   $\frac{dy}{3-4y}=dx$

3. 3

   Integrate both sides using the substitution $u = 3 - 4y$. Then $du=-4\,dy$, so $dy=-\tfrac14\,du$:

   $\int\frac{dy}{3-4y}=-\frac14\int\frac{du}{u}=-\frac14\ln|u|+C$

   The $-\tfrac14$ comes from the chain rule and is the step most often dropped:

   $-\frac14\ln|3-4y|=x+C$

4. 4

   Undo the logarithm. Multiply by $-4$ and exponentiate:

   $\ln|3-4y|=-4x+C_1\qquad\Longrightarrow\qquad |3-4y|=e^{C_1}e^{-4x}$

   Absorbing the sign of $3-4y$ into the constant lets the absolute value be dropped: $3-4y=Ae^{-4x}$ with $A$ any nonzero real.

5. 5

   Solve for y and fold the constants together.

   $y=\frac{3-Ae^{-4x}}{4}=\frac34-\frac{A}{4}e^{-4x}=\frac34+Ce^{-4x}$

   Relabelling $-\tfrac{A}{4}$ as $C$ is legitimate because $A$ was arbitrary, and allowing $C=0$ now recovers the equilibrium solution $y=\tfrac34$ that the division in step 2 excluded.

6. 6

   Verify by substitution. Differentiating, $y'=-4Ce^{-4x}$. Meanwhile $3-4y=3-3-4Ce^{-4x}=-4Ce^{-4x}$ — identical ✓. Numerically with $C=2.3$, at $x=0.1$ both sides equal $-6.16694$, and at $x=-0.4$ both equal $-45.5679$ ✓.

7. 7

   Interpret the behaviour. Since $e^{-4x}\to0$ as $x\to\infty$, every solution approaches $y=\tfrac34$ from whichever side it starts on: the equilibrium is stable, which is exactly what the negative coefficient $-4$ on $y$ predicts.