First-order differential equations practice questions — IB Maths AA

Factor the right-hand side: $S^{\mathrm{\prime }}=-{\displaystyle \frac{1}{5}}(S-20)$.

Separating the variables and integrating, $$\int \frac{dS}{S-20}=\int -\frac{1}{5}dt⟹\ln \mathrm{\mid }S-20\mathrm{\mid }=-\frac{t}{5}+k,$$ so $S-20=C{e}^{-t\mathrm{/}5}$.

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

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

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

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

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

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

Setting $y^{\mathrm{\prime }}=0$: $0=2y+6⟹y=-3$.

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

Verify $g$: $g^{\mathrm{\prime }}(x)=-1$ and $g+x=-x-1+x=-1$, so $g^{\mathrm{\prime }}=g+x$. ✓

The homogeneous equation $y^{\mathrm{\prime }}=y$ has general solution $C{e}^x$.

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

The constant solution comes from setting $y^{\mathrm{\prime }}=0$: $0=-2y+10$, so $y_e=5$.

Put $u=y-5$, so that $u^{\mathrm{\prime }}=y^{\mathrm{\prime }}$ and $y=u+5$: $$u^{\mathrm{\prime }}=-2(u+5)+10=-2u-10+10=-2u\mathrm{.}$$

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

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

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

Separate the variables in $y^{\mathrm{\prime }}=ky$: $$\frac{dy}{y}=kdt⟹\ln \mathrm{\mid }y\mathrm{\mid }=kt+c⟹y(t)=C{e}^{kt}\mathrm{.}$$

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

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

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

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

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

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

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