A linear equation has a whole family of solutions, one for each value of the constant, and an initial condition picks one. The order of work matters: find the general solution first, then use the condition to fix the constant — never fix a constant in the middle of a derivation. Afterwards check two things: that the answer satisfies both the equation and the condition, and on which interval it is valid.

Two tanks with identical plumbing, brine flowing in and mixture flowing out at the same rate. The equation describes the plumbing; how much salt each tank held at the start is a separate fact. The two follow different curves to the same final state.

Salt in a $10$-litre tank fed brine at $3$ grams per litre, $2$ litres a minute, and drained at the same rate: $A'=6-\tfrac{A}{5}$. Whether the tank starts with fresh water or saltier than the brine, the salt approaches $30$ grams — what ten litres of the incoming brine would hold.

To solve $y'+p(x)y=q(x)$ with $y(x_{0})=y_{0}$, find the general solution, substitute and , and solve for the constant. With definite integrals from the constant is built in. The solution exists and is unique on the largest interval containing on which and are continuous.

• For $y'+2y=4$ with $y(0)=5$: the general solution $$ gives , so and .

Find the general solution with its constant still free, then substitute the initial point and solve for the constant.

Salt arrives at the inflow rate times the incoming concentration, and leaves at the outflow rate times the concentration in the tank, which is the amount divided by the volume.

Great work! You now know how to:

• solve a linear initial value problem, general solution first and condition last
• write the solution with definite integrals so the condition is built in
• find the interval on which the solution is guaranteed to exist
• set up and solve a mixing problem