Substance A (initial amount a) decomposes into products P…
Substance $A$ (initial amount $a$) decomposes into products $P$ and $Q$, whose amounts $x(t),y(t)$ satisfy: $$ \frac{dx}{dt} = k_1(a-x-y),\qquad \frac{dy}{dt} = k_2(a-x-y) $$
Explanation of Method
Dividing the two rate equations shows $y$ is always proportional to $x$, fixing the ratio $k_2/k_1$ from the given data; adding the equations gives a separable ODE for the total amount decomposed, which determines the remaining rate constant.
- Divide the two equations to find the fixed ratio $y/x$.
- Use the data at $t=1$ to determine $k_2/k_1$.
- Add the equations to get a separable ODE for the total $x+y$.
- Use the data at $t=1$ again to fix the combined rate constant.
- Recover $x(t)$ and $y(t)$ individually.
Step 1 · Ratio of $k_1,k_2$
Dividing the equations, $dy/dx=k_2/k_1=\text{const}$, so $y=(k_2/k_1)x$ (using $x(0)=y(0)=0$). At $t=1$, $x=a/8$, $y=3a/8$, so
$$ \frac{k_2}{k_1}=\frac{3a/8}{a/8}=3\ \Longrightarrow\ k_2=3k_1=3k. $$
Step 2 · Total decomposition
Adding the equations, $\dfrac{d(x+y)}{dt}=4k(a-x-y)$. Separating and integrating with $x(0)+y(0)=0$,
$$ x(t)+y(t)=a\bigl(1-e^{-4kt}\bigr). $$
Step 3 · Determine $k$
At $t=1$: $x(1)+y(1)=a/8+3a/8=a/2$, so $1-e^{-4k}=1/2$, giving $e^{-4k}=1/2$, i.e. $e^{-4kt}=2^{-t}$.
Step 4 · Solve for $x(t)$ and $y(t)$
Since $y=3x$, $4x=a(1-2^{-t})$, so
$$ x(t)=\frac{a}{4}\bigl(1-2^{-t}\bigr),\qquad y(t)=\frac{3a}{4}\bigl(1-2^{-t}\bigr). $$
Step 5 · Verification
At $t=1$: $x(1)=\tfrac{a}{4}\cdot\tfrac12=\tfrac{a}{8}$ and $y(1)=\tfrac{3a}{4}\cdot\tfrac12=\tfrac{3a}{8}$, matching the given data, and both differential equations are satisfied by direct substitution. ✓
Summary Table
| Quantity | Value |
|---|---|
| Rate ratio | $k_2:k_1=3:1$ |
| Combined rate constant | $4k=\ln 2$ |
| Long-run split | $x\to a/4,\ y\to 3a/4$ as $t\to\infty$ |
Additional Notes
- The ratio $y/x=3$ holds at every instant, not just at $t=1$, since both amounts are driven by the same undecomposed reservoir $a-x-y$.
- As $t\to\infty$, all of $A$ decomposes ($A(t)=a\cdot2^{-t}\to0$), splitting in the fixed ratio $1:3$ between $P$ and $Q$.
- Verified against both given data points and by direct substitution into the original rate equations.
Result
The first integrals of the system are: