General and particular solutions

Solve $\dfrac{dy}{dx} = 2x$ by integrating, and you do not get one function — you get a whole family, $y = x^2 + C$ , one parabola for every value of $C$ . That family is the general solution. Left alone, a differential equation never pins down a single curve; it only fixes how $y$ changes, and integration always leaves behind an arbitrary constant of integration. To land on one specific curve out of the family, you need extra information — usually a point $(x_0, y_0)$ the curve must pass through, called an initial condition. Feeding that point into the general solution and solving for $C$ gives the particular solution, one exact curve. This distinction — family versus member, general versus particular — is the quiet backbone of the entire differential equations chapter: every method taught later produces a general solution first, and every initial condition supplied in a question is used exactly this way to finish the job.

The core insight

The general solution of a differential equation is a family of curves containing one arbitrary constant per order of the equation, and substituting a given initial condition into it solves for that constant to yield the one particular solution the question actually wants.

The core idea

A first-order differential equation $\dfrac{dy}{dx} = f(x,y)$ is solved by some process of integration, and integration always introduces a constant. So the outcome of "solving" a first-order equation is never a single function — it is a relation between $x$ , $y$ , and a constant $c$ .

General solution. A relation $g(x, y, c) = 0$ satisfying the differential equation, containing as many independent arbitrary constants as the order of the equation. A first-order equation's general solution has exactly one constant, usually written $c$ or $C$ ; a second-order equation's general solution has two. Geometrically, each value of the constant traces out one curve, and the general solution is the entire family of those curves sharing the same "slope rule."
Particular solution. One specific member of that family, singled out by assigning a numerical value to the constant. That value is found from an initial condition — a stated pair $(x_0, y_0)$ through which the required curve must pass — or, for higher-order equations, from as many conditions as there are constants.

For the first-order case this can be written compactly. If integrating $\dfrac{dy}{dx} = f(x)$ gives the general solution

y = F(x) + C,

where: $F(x)$ is any antiderivative of $f(x)$ and $C$ is the arbitrary constant of integration, then substituting a known pair $(x_0, y_0)$ fixes $C = y_0 - F(x_0)$ , and replacing $C$ with that number turns the family into the single required curve. The picture below makes the idea concrete: one differential equation, many parallel curves (the general solution), and one condition that selects exactly one of them.

Figure

A family of parabolas y equals x squared plus C for several values of C shown as parallel curves with one curve highlighted after it is pinned down by a point at x equals one y equals three

Derivation

Derivation AFinding the particular solution from an initial condition

The goal is to show, as a general four-step procedure that works for any first-order equation solvable by direct integration, exactly how a stated initial condition converts the general solution's arbitrary constant into one fixed number, so that the particular solution can be written down and independently checked by differentiation.

Step 1. Solve the differential equation to obtain the general solution. Given $\dfrac{dy}{dx} = f(x)$ , integrate both sides: $\displaystyle\int dy = \int f(x)\,dx$ , which gives $y = F(x) + C$ where $F(x)$ is an antiderivative of $f(x)$ and $C$ is the arbitrary constant. At this stage $C$ is unknown, so the equation describes an entire family of curves, not one curve.

Step 2. Substitute the given initial condition $(x_0, y_0)$ into the general solution. Since the required curve must pass through this point, its coordinates must satisfy the relation: $y_0 = F(x_0) + C$ . This produces a single ordinary equation in the one unknown $C$ , because $x_0$ and $y_0$ are given numbers.

Step 3. Solve that equation for $C$ . Rearranging gives $C = y_0 - F(x_0)$ , a specific number rather than an unknown symbol. This is the step that "uses up" the extra information in the problem and removes the ambiguity of the family.

Step 4. Substitute this numerical value of $C$ back into the general solution from Step 1 to write the particular solution $y = F(x) + \big(y_0 - F(x_0)\big)$ . As a check, differentiate this result: it must return the original $f(x)$ , and substituting $x = x_0$ must return $y_0$ ; both checks confirm the particular solution is correct. Verified: for $f(x) = 2x$ , $(x_0,y_0) = (1,3)$ , Step 1 gives $y = x^2 + C$ , Step 2 gives $3 = 1 + C$ , Step 3 gives $C = 2$ , Step 4 gives $y = x^2 + 2$ , and differentiating recovers $dy/dx = 2x$ with $y(1) = 1 + 2 = 3$ as required.

Worked example

Find the particular solution of $\dfrac{dy}{dx} = 3x^2$ , given that $y = 5$ when $x = 0$ .

Step 1. Find the general solution by integrating both sides. $dy = 3x^2\,dx \implies \displaystyle\int dy = \int 3x^2\,dx \implies y = x^3 + C.$

Step 2. Apply the initial condition $x = 0,\ y = 5$ . $5 = (0)^3 + C \implies C = 5.$

Step 3. Substitute $C = 5$ back into the general solution to get the particular solution. $y = x^3 + 5.$

Step 4. Check by differentiating. $\dfrac{dy}{dx} = 3x^2$ , matching the original equation, and at $x = 0$ , $y = 0 + 5 = 5$ as required. The particular solution is $y = x^3 + 5$ .

Competitive stretch

The same four steps extend to a second-order case even though solving second-order equations is outside the CBSE syllabus. If the general solution is given as $y = A\sin x + B\cos x$ with $y(0) = 2$ and $\dfrac{dy}{dx}\big|_{x=0} = 1$ , the first condition gives $B = 2$ directly (since $\sin 0 = 0,\ \cos 0 = 1$ ), and differentiating to get $\dfrac{dy}{dx} = A\cos x - B\sin x$ then applying the second condition gives $A = 1$ . The particular solution is $y = \sin x + 2\cos x$ . Two constants needed two conditions — exactly matching the rule that the number of arbitrary constants equals the order of the equation.

Common traps

Stopping at the general solution. If the question supplies an initial condition, the expected final answer is the particular solution with $C$ evaluated, not the family left with $C$ symbolic.
Substituting the initial condition into the differential equation itself. The condition belongs in the general solution (the relation with $C$ ), applied after integrating — not into $\dfrac{dy}{dx} = f(x,y)$ .
Swapping $x_0$ and $y_0$ . Carelessly putting the $y$ -value where $x$ should go (or vice versa) when substituting into the general solution gives the wrong constant and a wrong final curve.
Dropping the "+C" while integrating. Skipping the constant of integration on simple polynomial or trigonometric antiderivatives makes it impossible to apply the initial condition correctly afterward.

Ask the tutor: if turning a general solution into a particular solution ever feels shaky, ask your AI tutor to walk through the four-step procedure on a fresh differential equation of your own choosing.

Why does integrating a differential equation always leave behind a constant, and why does that matter?

Integration is the reverse of differentiation, and differentiation destroys any constant term because its derivative is zero. So when you integrate to undo a derivative, you cannot know what constant was originally there — it could have been anything, and every choice still satisfies the original differential equation. That is exactly why the "solution" of a differential equation is naturally a family, not a single function, and why extra information (an initial condition) is needed to pick out one member of that family.

What if I am given two conditions instead of one — does that mean I made a mistake?

Not necessarily; it depends on the order of the equation. A first-order equation needs exactly one condition to fix its one constant. A second-order equation's general solution has two constants, so it genuinely needs two conditions, applied one to the solution itself and one to its derivative, as in the competitive stretch example above. CBSE Class 12 mainly works with first-order equations, so one condition is the norm, but recognizing why higher orders need more conditions is useful conceptual understanding.

Is the particular solution "more correct" than the general solution?

No — both are equally valid solutions of the differential equation; they differ only in how much information they carry. The general solution is more informative in one sense (it describes every possible curve satisfying the equation), while the particular solution answers a narrower, more specific question (which one curve satisfies both the equation and a given starting point). A question that supplies an initial condition is asking for the particular solution specifically, so leaving the constant unevaluated there would be an incomplete answer, not an equally valid alternative.