Differential equations — practice questions

This equation is in the form $\frac{dy}{dx} = f(x)g(y)$, so we can separate the variables.

Step 1: Separate the variables. $\frac{1}{y+1} dy = \frac{1}{x^2} dx$

Step 2: Integrate both sides. $\int \frac{1}{y+1} dy = \int x^{-2} dx$

Step 3: Use the boundary condition to find c. We are given $y=1$ when $x=1$. $\ln|1+1| = -\frac{1}{1} + c$ $\ln(2) = -1 + c$ $c = \ln(2) + 1$

Step 4: Substitute c back and solve for y. $\ln|y+1| = -\frac{1}{x} + 1 + \ln(2)$ To make $y$ the subject, we exponentiate both sides.

The equation is in the linear form $dN/dt + P(t)N = Q(t)$ with $P(t) = 0.2$ and $Q(t) = 100e^{-0.1t}$.

Step 1: Find the integrating factor (I.F.). $P(t) = 0.2$, so $\int P(t) dt = \int 0.2 dt = 0.2t$. $I.F. = e^{0.2t}$

Step 2: Multiply the entire DE by the I.F. $e^{0.2t} \frac{dN}{dt} + 0.2e^{0.2t}N = 100e^{-0.1t}e^{0.2t}$

Step 3: The left side becomes $d/dt (N \cdot I.F.)$. $\frac{d}{dt}(N e^{0.2t}) = 100e^{0.1t}$

Step 4: Integrate both sides with respect to t. $\int \frac{d}{dt}(N e^{0.2t}) dt = \int 100e^{0.1t} dt$ $N e^{0.2t} = \frac{100}{0.1}e^{0.1t} + c$ $N e^{0.2t} = 1000e^{0.1t} + c$

Step 5: Use the initial condition to find c. Given $N=500$ when $t=0$. $500 e^{0.2(0)} = 1000e^{0.1(0)} + c$ $500 \cdot 1 = 1000 \cdot 1 + c$ $500 = 1000 + c \implies c = -500$

Step 6: Substitute c and solve for N. $N e^{0.2t} = 1000e^{0.1t} - 500$ $N = \frac{1000e^{0.1t} - 500}{e^{0.2t}}$ $N = 1000e^{-0.1t} - 500e^{-0.2t}$