A function y = f(x) satisfies the differential equation dy/dx = 3x^2 (y - 4). If f(0) = 6, find the particular solution.

Step-by-step solution

1. 1

   Spot that the equation is separable and check the special solution. The right-hand side is a function of $x$ times a function of $y$, so the variables can be split. Before dividing, note that $y=4$ makes the right side zero and is itself a constant solution; the initial value $f(0)=6\neq 4$ tells us our solution is not that one, so dividing by $y-4$ is legitimate near the initial point.

2. 2

   Separate the variables. Move all $y$ to the left and all $x$ to the right:

   $\frac{1}{y-4}\,dy=3x^{2}\,dx$

3. 3

   Integrate both sides. The left is a logarithm, the right a power:

   $\int\frac{1}{y-4}\,dy=\int 3x^{2}\,dx\ \Longrightarrow\ \ln|y-4|=x^{3}+C$

   Only one constant is needed because the two sides' constants can be merged.

4. 4

   Exponentiate to solve for $y$. Applying $e^{(\cdot)}$ to both sides,

   $|y-4|=e^{x^{3}+C}=e^{C}e^{x^{3}}$

   Since $e^{C}$ is an arbitrary positive constant and the sign of $y-4$ is absorbed by allowing the constant to be negative, write

   $y-4=Ke^{x^{3}}\ \Longrightarrow\ y=4+Ke^{x^{3}}$

   with $K$ any real constant ($K=0$ recovers the special solution $y=4$).

5. 5

   Apply the initial condition. Substituting $x=0$, $y=6$ and using $e^{0}=1$:

   $6=4+K\cdot 1\ \Longrightarrow\ K=2$

   So the particular solution is

   $y=4+2e^{x^{3}}$

6. 6

   Verify the solution satisfies both requirements. Differentiating, $\dfrac{dy}{dx}=2\cdot 3x^{2}e^{x^{3}}=6x^{2}e^{x^{3}}$, while $3x^{2}(y-4)=3x^{2}\cdot 2e^{x^{3}}=6x^{2}e^{x^{3}}$ — identical. A central-difference check at $x=0.5$, $1.1$ and $-0.8$ matches $3x^{2}(y-4)$ to five decimal places, and $y(0)=4+2=6$ as required.