First-order differential equations practice questions — IB Maths AA

Factor the right-hand side: $S' = -\dfrac{1}{5}\left(S - 20\right)$.

Separating the variables and integrating, $$\int \frac{dS}{S - 20} = \int -\frac{1}{5}\,dt \quad \Longrightarrow \quad \ln|S - 20| = -\frac{t}{5} + k,$$ so $S - 20 = Ce^{-t/5}$.

The initial condition gives $100 - 20 = C$, hence $C = 80$ and $$S(t) = 20 + 80e^{-t/5}.$$

Now impose $S(t) = 40$: $$80e^{-t/5} = 20 \quad \Longrightarrow \quad e^{-t/5} = \frac{1}{4} \quad \Longrightarrow \quad \frac{t}{5} = \ln 4.$$

So $t = 5\ln 4$, and since $\ln 4 = 2\ln 2$, $$t = 10\ln 2.$$

General solution of $y' = 3y$: $y(x) = C e^{3x}$.

Apply $y(0) = 5$: $C e^{0} = C = 5$.

So $y(x) = 5 e^{3x}$.

Setting $y' = 0$: $0 = 2y + 6 \implies y = -3$.

Equivalently, $y_p = -\dfrac{b}{a} = -\dfrac{6}{2} = -3$.

Verify $g$: $g'(x) = -1$ and $g + x = -x - 1 + x = -1$, so $g' = g + x$. ✓

The homogeneous equation $y' = y$ has general solution $Ce^{x}$.

So the general solution of $y' = y + x$ is $y(x) = -x - 1 + C e^{x}$.

The constant solution comes from setting $y' = 0$: $0 = -2y + 10$, so $y_e = 5$.

Put $u = y - 5$, so that $u' = y'$ and $y = u + 5$: $$u' = -2(u + 5) + 10 = -2u - 10 + 10 = -2u.$$

A function whose derivative is $-2$ times itself is a multiple of $e^{-2x}$, so $u(x) = Ce^{-2x}$ and therefore $$y(x) = 5 + Ce^{-2x}.$$

The initial condition $y(0) = 1$ gives $5 + C = 1$, hence $C = -4$.

Check: $y' = 8e^{-2x}$, while $-2y + 10 = -2\left(5 - 4e^{-2x}\right) + 10 = 8e^{-2x}$, and $y(0) = 5 - 4 = 1$. $$y(x) = 5 - 4e^{-2x}$$

Separate the variables in $y' = ky$: $$\frac{dy}{y} = k\,dt \implies \ln|y| = kt + c \implies y(t) = Ce^{kt}.$$

The condition $y(0) = 12$ gives $Ce^{0} = C = 12$, so $y(t) = 12e^{kt}$.

The condition $y(2) = 3$ pins the rate: $$12e^{2k} = 3 \implies e^{2k} = \frac{1}{4}.$$

There is no need to extract $k$ itself, because $e^{4k}$ is the square of $e^{2k}$: $$y(4) = 12e^{4k} = 12\left(e^{2k}\right)^{2} = 12\cdot\frac{1}{16}.$$

Hence $$y(4) = \frac{3}{4}.$$

Differentiate the given function: $$y'(x) = 4\cdot(-3)e^{-3x} = -12e^{-3x}.$$

To compare $y'$ with $y$, factor the given function back out of this expression: $$-12e^{-3x} = -3\cdot\left(4e^{-3x}\right) = -3\,y(x).$$

The amplitude $4$ appears in both $y$ and $y'$, so it cancels in the ratio $y'/y$ and never enters the rate; only the coefficient of $x$ in the exponent survives: $$y' = -3y.$$