From a General Solution to a Model
A separable differential equation usually produces a family of solutions, not one particular function. The arbitrary constant in that family represents the fact that many different solution curves can satisfy the same differential equation. An initial condition specifies one point on the curve and lets us determine which member of the family models the situation.
For an initial value problem, the differential equation describes how a quantity changes, and the initial condition gives its value at a known input. The goal is to use both pieces of information: first determine the particular solution, then evaluate it at the requested input and explain the result in context.
How the Initial Condition Determines the Constant
Consider a separable differential equation in the form \(y'=f(x)g(y)\). Where \(g(y)\neq 0\), separate the variables and integrate:
Let \(G\) be an antiderivative of \(1/g\), and let \(F\) be an antiderivative of \(f\). The integrated equation is
If the initial condition is \(y(x_0)=y_0\), substitute \(x_0\) for \(x\) and \(y_0\) for \(y\):
This is why the initial condition determines the constant: it gives a specific equation involving \(C\). The equation may be left in implicit form, or it may be solved for \(y\). If solving requires a square root or another inverse operation, choose the branch that agrees with \(y(x_0)=y_0\).
Why This Procedure Works
The constant is not chosen by guesswork. It records the difference between the two antiderivative values at a solution point. The following argument justifies the procedure and explains why the initial value selects a particular curve, at least locally when the equation can be separated and inverted.
Assume \(g(y)\neq 0\) along the part of the solution being considered. Dividing by \(g(y)\) gives \(y'/g(y)=f(x)\).
By the chain rule, \(\frac{d}{dx}G(y(x))=G'(y)y'=\frac{y'}{g(y)}\), because \(G'(y)=1/g(y)\).
Since \(\frac{d}{dx}F(x)=f(x)\), the two derivatives are equal, so \(G(y(x))-F(x)\) is constant on the interval. Thus \(G(y)=F(x)+C\).
At \(x=x_0\), the solution has \(y=y_0\), so \(C=G(y_0)-F(x_0)\). If \(G\) is invertible on the relevant range, this relation determines \(y\) there.
This argument also gives a useful check: differentiating the implicit equation \(G(y)=F(x)+C\) recovers the separated differential equation, as long as \(g(y)\neq 0\).
Worked Examples
Worked Example: Determine the Constant and Evaluate the Solution
Solve the initial value problem \(y'=\frac{x}{y+1}\), \(y(0)=1\), and find \(y(3)\).
Step 1: Separate variables. The equation is defined only when \(y\neq -1\). Multiply by \(y+1\) and by \(dx\):
Step 2: Integrate both sides. Keep the full antiderivative of \(y+1\):
Step 3: Use the initial condition. Substitute \(x=0\) and \(y=1\):
Step 4: Solve for \(y\) and choose the branch. Multiply the implicit equation by \(2\), then complete the square:
Therefore \(y+1=\pm\sqrt{x^2+4}\). The initial value gives \(y(0)+1=2>0\), so the matching branch is
Step 5: Evaluate.
Check. At the initial input, \(y(0)=\sqrt{4}-1=1\), as required. Differentiating the solution gives \(y'=x/\sqrt{x^2+4}\). Also \(y+1=\sqrt{x^2+4}\), so the right side of the differential equation is \(x/(y+1)=x/\sqrt{x^2+4}\). The derivative agrees with the equation. Thus the requested value is \(y(3)=\sqrt{13}-1\), approximately \(2.61\) to two decimal places.
Worked Example: Keep the Initial Value on the Correct Side
Solve \(y'=\frac{1+x}{2y}\), \(y(0)=2\), and evaluate the solution at \(x=2\).
Step 1: Separate variables. Since the differential equation is undefined at \(y=0\), separate on an interval where \(y\neq 0\):
Step 2: Integrate.
Step 3: Determine \(C\). Substitute \(x=0\) and \(y=2\):
The implicit solution is \(y^2=x+\frac{x^2}{2}+4\). Solving gives two possible signs, but \(y(0)=2\) is positive, so use the positive branch:
Step 4: Evaluate at \(x=2\).
Check. The initial condition holds because \(y(0)=\sqrt{4}=2\). Differentiating \(y^2=x+\frac{x^2}{2}+4\) gives \(2yy'=1+x\), hence \(y'=(1+x)/(2y)\), exactly the given equation. Therefore \(y(2)=2\sqrt{2}\), or approximately \(2.83\) to two decimal places.
Worked Example: Evaluate and Interpret a Normalized Model
A normalized quantity \(y\) satisfies \(\frac{dy}{dt}=\frac{3t^2}{y^2}\) and \(y(0)=1\). Find \(y(1)\) and interpret it as a model value.
Step 1: Separate variables. Since the initial value is positive and the equation is undefined at \(y=0\), work on the positive branch:
Step 2: Integrate and apply the initial condition.
Multiplying by \(3\) gives \(y^3=3t^3+1\). The initial value selects the positive cube root:
Step 3: Evaluate at \(t=1\).
Check. The initial condition holds because \(y(0)=\sqrt[3]{1}=1\). Differentiating \(y^3=3t^3+1\) gives \(3y^2y'=9t^2\), so \(y'=3t^2/y^2\), as required. The model predicts a normalized quantity of \(\sqrt[3]{4}\), approximately \(1.59\) to two decimal places, at the input \(t=1\).
Common Pitfall: Substituting Incorrectly for the Constant
A frequent error is to substitute the initial values into the integrated equation but forget to evaluate the entire left side. In the first example, the integrated equation was \(\frac{y^2}{2}+y=\frac{x^2}{2}+C\). At \((x,y)=(0,1)\), the left side is \(\frac{1}{2}+1=\frac{3}{2}\), so \(C=\frac{3}{2}\).
A wrong calculation might set \(C=1\), using only the initial value \(y(0)=1\) and overlooking the other term in the antiderivative. That would give \((y+1)^2=x^2+3\), and then the positive branch would predict \(y(0)=\sqrt{3}-1\), which is not \(1\). Substituting the proposed solution back into the initial condition exposes the error immediately.
Reporting the Requested Value
On a free-response question, a correct numerical or symbolic value is not always a complete interpretation. Identify which input was used and what the output represents. If the model describes a real quantity with units, include the units in the final statement. If rounding is requested, keep exact values through the calculation and round only at the end, stating the precision.
Also distinguish the value predicted by the particular solution from the general family of solutions. The differential equation alone describes a family; the initial condition selects the member used for the requested evaluation. A complete response should make that link clear.
Check Your Understanding
For each question, focus on how the initial value selects and checks a particular solution.
- For \(y'=x/(y+2)\) with \(y(0)=1\), what equation determines the integration constant after separating and integrating?
- Why must a square-root solution be checked against the initial value before choosing its sign?
- In \(y'=f(x)g(y)\), what possibility can be lost by dividing by \(g(y)\)?
- For \(y^2=x^2+5\) and the condition \(y(0)=\sqrt{5}\), which branch is appropriate?
- Name two checks that help verify a particular solution to an initial value problem.
