From a General Solution to a Particular Solution
A differential equation can have many solutions. A general solution describes a family of curves, often using an arbitrary constant introduced during integration. An initial condition specifies the value of the dependent variable at a particular input. When the given point is allowed by the equation and the solution, it can select a curve from that family.
For instance, a general solution might contain a constant \(C\). Substituting the given input and output into the solution determines \(C\). That calculation is only the beginning: the resulting curve must still satisfy the differential equation, pass through the specified point, and be defined there.
In this tutorial, the general solutions come from separating variables and integrating. Once such a family has been found, the initial condition determines its constant when the substitution is valid. The initial condition may also determine which branch of an implicit relation to use.
How the Initial Condition Determines the Constant
Suppose integration gives a family of solutions in the form \(F(x,y)=C\). If a particular solution passes through \((x_0,y_0)\), then substituting that point gives \(C=F(x_0,y_0)\). This is the basic reason initial conditions select members of a family.
After finding \(C\), keep track of any restrictions made while separating variables. For example, dividing by \(y\) requires \(y\ne 0\), and an equation containing \(\sqrt{y}\) in a denominator requires \(y>0\). Such restrictions are not optional: they determine where the original differential equation is defined.
If the integrated relation is implicit, the initial point can also distinguish between branches. Solving an equation such as \(y^2=G(x)\) produces positive and negative possibilities wherever \(G(x)>0\); the initial value tells which one can pass through the specified point.
Worked Example: An Implicit Relation Selects a Branch
Worked Example: An Implicit Relation Selects a Branch
Solve the initial-value problem \(y'=\frac{x}{y}\), \(y(1)=2\).
The right side requires \(y\ne 0\), so separation is valid only where \(y\ne 0\). Separate variables and integrate:
Multiply the relation by \(2\), absorbing the factor \(2\) into the arbitrary constant:
Use \(x=1\) and \(y=2\) to determine the constant:
The implicit particular solution is \(y^2-x^2=3\). To express it explicitly, solve for \(y\):
At \(x=1\), the initial value is \(y=2\), so the positive branch is required. The particular solution is \(y=\sqrt{x^2+3}\). Since \(x^2+3>0\) for every real \(x\), this branch is defined for all real \(x\), and it never equals zero.
Check the initial condition:
Check the differential equation by differentiating the explicit function:
Thus the selected branch satisfies both the differential equation and the initial condition on its domain.
Check the Domain Before Substituting
Finding a value for \(C\) by algebra does not prove that an initial-value problem has a solution. The differential equation must be defined at the initial point, and the candidate solution must be defined there too. A substitution that uses a point outside the domain can produce a constant without producing a valid particular solution.
For example, if the equation contains \(1/\sqrt{y}\), then its right side is defined only for \(y>0\). An initial condition with \(y_0=0\) is therefore not allowed, even if a formula obtained after integration appears to have a value at that point. Always check the original equation’s restrictions, not just the simplified solution formula.
Worked Example: Integrating Carefully and Respecting a Restricted Domain
Worked Example: Integrating Carefully and Respecting a Restricted Domain
Solve \(y'=\frac{1}{2\sqrt{y}}\), \(y(0)=1\).
The right side is defined only when \(y>0\). In this region, separate variables by multiplying both sides by \(2\sqrt{y}\,dx\):
Integrate with respect to the corresponding variables. Since \(\sqrt{y}=y^{1/2}\), its integral is \(\frac{2}{3}y^{3/2}\); the factor \(2\) in the integrand must also be included:
Thus the integrated family is
Apply \(y(0)=1\):
Now solve for \(y\):
The original equation requires \(y>0\), so the expression inside the power must be positive. Therefore \(1+\frac{3x}{4}>0\), or \(x>-\frac{4}{3}\). This interval contains the initial input \(x=0\), and the solution is positive there.
Check the initial condition:
For the differential-equation check, differentiate the function:
Also, \(\sqrt{y}=\left(1+\frac{3x}{4}\right)^{1/3}\). Consequently,
The function satisfies the initial condition and the differential equation for \(x>-\frac{4}{3}\). In particular, it is a valid solution on an interval containing \(x=0\).
Why Substitution Works
The constant-finding step follows directly from what a solution means. Suppose the integrated family is \(F(x,y)=C\), and a particular solution \(y(x)\) satisfies this relation. If it also passes through \((x_0,y_0)\), then \(y(x_0)=y_0\). Substituting \(x=x_0\) into the relation gives \(F(x_0,y(x_0))=C\), and using the initial condition gives \(F(x_0,y_0)=C\). Thus there is no freedom left in the constant for that specified point.
This argument explains how to find a candidate constant. It does not by itself prove that the candidate relation defines a valid solution. That still requires a domain check and a verification against the differential equation. In an implicit relation, it may also be necessary to identify which branch passes through the initial point.
Worked Example: An Implicit Solution That Is Best Left Implicit
Worked Example: An Implicit Solution That Is Best Left Implicit
Solve \(y'=\frac{\cos x}{1+y^2}\), \(y(0)=0\), giving the solution implicitly.
The denominator \(1+y^2\) is positive for every real \(y\), so the differential equation is defined for all real pairs \((x,y)\). Separate variables and integrate:
Use the initial condition to find the constant:
The particular solution is described implicitly by \(y+\frac{y^3}{3}=\sin x\). This relation gives one real \(y\) for each real \(x\): the function \(H(y)=y+\frac{y^3}{3}\) has derivative \(H'(y)=1+y^2>0\), so it is strictly increasing, and its values extend without bound in both directions. Hence its range is all real numbers. Moreover, \(H'(y)\ne 0\), so implicit differentiation is valid.
Differentiate the relation with respect to \(x\):
At \(x=0\), \(y=0\) makes the implicit relation read \(0+0=\sin(0)\), so the initial condition holds. Both the differential equation and this implicit solution are defined there; indeed, the relation defines the solution for every real \(x\).
A Common Mistake: Accepting an Invalid Initial Point
Consider \(y'=\frac{1}{y}\) with the proposed initial condition \(y(0)=0\). Separation requires division by \(1/y\), or equivalently multiplication by \(y\); either way, the original equation is defined only for \(y\ne 0\). The separated and integrated relation on that restricted domain is:
A mistaken approach substitutes \(x=0\), \(y=0\), obtains \(C=0\), and then writes \(y=\sqrt{2x}\). The algebraic formula is real for \(x\ge 0\), but this does not make it a solution of the original initial-value problem. At the initial point, the right side \(1/y\) is undefined. In addition, the derivative of \(\sqrt{2x}\) is \(1/\sqrt{2x}\) for \(x>0\), which is undefined at \(x=0\).
The correction is to check the original equation before accepting the result: \(y_0=0\) is outside its domain, so this proposed initial condition cannot be satisfied by a solution of the differential equation. Substitution into an integrated formula cannot override the domain restriction used to obtain that formula.
Check Your Understanding
For each problem, determine the constant or particular solution as requested, and check whether the initial point lies in the relevant domain.
- For \(y'=x/y\), the implicit family is \(y^2-x^2=C\). Use \(y(2)=3\) to find \(C\), then identify the branch through the initial point.
- For \(y'=\frac{1}{2\sqrt y}\), explain why an initial condition with \(y(0)=0\) is not allowed, even if a separated relation can be manipulated algebraically.
- For \(y'=\frac{\cos x}{1+y^2}\), use the implicit family \(y+\frac{y^3}{3}=\sin x+C\) and \(y(0)=1\) to find \(C\).
- Why must a solution of an initial-value problem be defined on an interval containing its initial input?
- After finding a candidate particular solution, name two checks that help establish it is valid.
