General and Particular Solutions: Definitions, Examples

What Are General And Particular Solutions?

General and particular solutions are the two ways of writing the answer to a differential equation. The general solution contains arbitrary constants and describes every solution in one formula, while a particular solution is a single function obtained by giving those constants specific values, usually from initial conditions. For example, $y = x^2 + C$ is the general solution of $\dfrac{dy}{dx} = 2x$, and $y = x^2 + 3$ is the particular solution that passes through the point $(2, 7)$.

A differential equation is an equation that links an unknown function to its derivatives (the full classification lives on the ordinary differential equations page). For an equation of order $n$, the general solution has the shape

$$y = \phi(x, C_1, C_2, \ldots, C_n),$$

where:

• $x$ is the independent variable and $y$ is the unknown function,

• $C_1, C_2, \ldots, C_n$ are arbitrary constants, each free to take any real value,

• $n$ is the order of the equation, the highest derivative that appears.

A particular solution is the same expression with every $C_k$ replaced by a number. The condition that makes this work is that the constants are essential: none of them can be merged into another. Writing $y = x + C_1 + C_2$ looks like it has two constants, but $C_1 + C_2$ is really one constant, so that expression counts as having only one.

Why Does A Differential Equation Have Infinitely Many Solutions?

A differential equation describes how a quantity changes, not where it starts, so infinitely many functions share the same rule of change. Differentiation throws away constant information: $x^2$, $x^2 + 3$ and $x^2 - 100$ all have derivative $2x$. Undoing a derivative therefore always brings back a free constant, the same constant of integration you meet with every indefinite integral.

The physics reader on Stack Exchange who asked "What's the meaning of the general solution and the particular solution in differential equations?" was puzzled that a single equation could have "a whole host" of solutions. The answer is that the equation encodes the law, and the law is shared:

• The law is the general solution. Newton's second law says a thrown ball has acceleration $-9.8,\text{m/s}^2$, whoever throws it.

• The situation is the particular solution. The height and speed at the moment of release are different for every throw, and they fix the constants.

• Prediction needs both. Without the starting data you know the shape of the arc; with it, you know where the ball lands.

This is why scientists and engineers write the general solution first. A model of a falling object or a cooling engine is solved once, and every new set of measured starting values gives a new particular solution of the same general one. The OpenStax treatment of differential equations sets out the same split between a family of solutions and the one member that fits an initial value.

Why Does The Number Of Constants Match The Order?

The general solution of an $n$th-order differential equation contains exactly $n$ essential arbitrary constants, because solving it takes $n$ integrations and each one adds a constant. A first-order equation gives one constant, a second-order equation gives two, and so on. The order and degree of a differential equation page explains how to read the order off an equation; here the point is what the order costs you.

The cleanest way to see it is in reverse. Start from a family of curves and eliminate the constants by differentiating: you need one derivative per constant, so the differential equation you land on has order equal to the number of constants.

One constant gives a first-order equation. Take the family $y = Cx^2$:

$$\frac{dy}{dx} = 2Cx$$

$$C = \frac{y}{x^2}$$

$$\frac{dy}{dx} = \frac{2y}{x}$$

Two constants give a second-order equation. Take $y = A\sin x + B\cos x$:

$$\frac{dy}{dx} = A\cos x - B\sin x$$

$$\frac{d^2y}{dx^2} = -A\sin x - B\cos x = -y$$

$$\frac{d^2y}{dx^2} + y = 0$$

A Math Stack Exchange user asked, in a thread titled "Forming an O.D.E for $y=A\sin(Bx)+B\cos(Bx)$", why the constants refused to disappear. In that family the letter $B$ sits inside the sine and cosine as well as in front of the cosine, so it is not a simple multiplier and the clean two-step elimination above does not apply. Constants that enter linearly, as $A$ and $B$ do in $A\sin x + B\cos x$, are the ones that eliminate neatly.

How Do You Find A Particular Solution?

You find a particular solution by solving for the general solution first, then substituting the given conditions to fix every constant. That is also the short answer to a Quora question, "What is the particular solution of the ordinary differential equation?": it is the one member of the general family that meets the stated conditions. The conditions usually arrive as an initial condition such as $y(0) = 4$, and a differential equation paired with its initial conditions is called an initial value problem.

1. Solve the equation in general form. Integrate, separate or apply the right method, and keep every constant visible.

2. Count the conditions. You need as many independent conditions as there are constants. One constant needs one condition; two need two (often a value and a slope).

3. Substitute each condition. Replace $x$ and $y$ (or $y'$) by the given numbers to get equations in the constants.

4. Solve for the constants. One condition gives one equation; two conditions give a small linear system.

5. Write the particular solution in full. Put the numbers back into the general solution and state the final function.

6. Check it. Substitute the function into the original equation and into each condition.

Step 5 matters more than it looks. Finding $C = 6$ is not the answer; $y = x^2 + 6$ is.

How Do You Check That A Function Is A Solution?

You check a proposed solution by substituting it, and its derivatives, into the differential equation and confirming both sides agree for every $x$ in the interval. To confirm that $y = C_1 e^{2x} + C_2 e^{-3x}$ solves $y'' + y' - 6y = 0$, differentiate twice:

$$y' = 2C_1 e^{2x} - 3C_2 e^{-3x}$$

$$y'' = 4C_1 e^{2x} + 9C_2 e^{-3x}$$

$$y'' + y' - 6y = (4 + 2 - 6)C_1 e^{2x} + (9 - 3 - 6)C_2 e^{-3x} = 0$$

Both brackets vanish for every choice of $C_1$ and $C_2$, so the whole two-parameter family satisfies the equation. This check is also the fastest way to catch an algebra error before it reaches the final answer, which is why every example below ends with one.

What Is The Difference Between A General Solution And A Particular Solution?

The difference is the constants: a general solution still contains arbitrary constants and represents a family of curves, while a particular solution contains none and represents one curve. A Wyzant question puts it exactly this way, asking for the difference between a general solution and a particular solution when solving a differential equation. The table compares the two side by side, together with the singular solution covered in the next section.

The terms shift meaning slightly in second-order work, which is the confusion behind the Math Stack Exchange question "Confusion with Regards to General and Particular Solution Terminology in Differential Equations". That case gets its own section further down.

What Is A Singular Solution?

A singular solution is a solution of a differential equation that cannot be obtained from the general solution by any choice of the arbitrary constants. It usually appears when a solving step divides by an expression that can be zero, so the step quietly discards a valid solution.

A Math Stack Exchange thread, "Clarification on the definition of General Solution", uses exactly this example:

$$\frac{dy}{dx} = 3y^{2/3}$$

Separating the variables divides by $y^{2/3}$, which assumes $y \neq 0$, and gives the family $y = (x + C)^3$. But the constant function $y = 0$ also satisfies the equation, since its derivative is $0$ and $3 \cdot 0^{2/3} = 0$. No value of $C$ turns $(x + C)^3$ into the zero function, so $y = 0$ is a singular solution. It even shares the point $(2, 0)$ with the particular solution $y = (x - 2)^3$, so two different solutions pass through one point.

The classic source is the Clairaut equation $y = xy' + (y')^2$. Its general solution is the family of straight lines $y = cx + c^2$, and its singular solution is the parabola $y = -\dfrac{x^2}{4}$, which touches every one of those lines (the parabola is their envelope). Alexis Clairaut studied these equations in 1734 and gave a singular solution alongside the general one.

Singular solutions are the main reason some textbooks call the general solution "the solution with $n$ arbitrary constants" rather than "every solution". The separable differential equations page shows how to catch the lost solutions while separating.

Does Particular Solution Mean Something Different For Second-Order Equations?

Yes. For linear second-order equations with a nonzero right side, "a particular solution" usually means any one solution $y_p$ of the full equation, with no constants, and the general solution is $y = y_c + y_p$, where $y_c$ (the complementary function) solves the matching equation with zero on the right. That use is different from "the particular solution of an initial value problem", and mixing the two causes the confusion in the Math Stack Exchange threads "Difference between CF particular solution general solution" and "Why is the particular solution of an non-homogeneous differential equation not the full (general) solution?".

Take $y'' + y = x$:

$$y_c = C_1\cos x + C_2\sin x$$

$$y_p = x$$

$$y = x + C_1\cos x + C_2\sin x$$

Here $y_p = x$ is a particular solution in the first sense: it satisfies the equation but ignores any starting data. Add the conditions $y(0) = 0$ and $y'(0) = 0$:

$$y(0) = C_1 = 0$$

$$y'(0) = 1 + C_2 = 0, \quad C_2 = -1$$

$$y = x - \sin x$$

Now $y = x - \sin x$ is the particular solution in the second sense. Both are correct uses of the word; the question to ask is "particular to what?". A Physics Forums thread with the title "General solution vs particular solution" makes the same point: there is no single "the" particular solution $y_p$, because adding any piece of $y_c$ to it gives another valid one.

Does Every Set Of Conditions Give Exactly One Particular Solution?

No. Initial conditions given at a single point usually produce exactly one particular solution, but conditions split across two points (boundary conditions) can produce none or infinitely many. The general solution of $y'' + y = 0$ is $y = C_1\cos x + C_2\sin x$.

• Conditions $y(0) = 0$ and $y(\pi) = 0$: the first forces $C_1 = 0$, and the second gives $C_2 \sin\pi = 0$, true for every $C_2$. Infinitely many particular solutions $y = C_2 \sin x$.

• Conditions $y(0) = 0$ and $y(\pi) = 1$: $C_1 = 0$ again, but then $y(\pi) = 0 \neq 1$. No particular solution at all.

When uniqueness is guaranteed for initial value problems is a theorem of its own (Picard and Lindelöf), covered on the initial value problems page. For the boundary case, the lesson is simply to solve the system for the constants and read what it says.

Which Methods Produce The General Solution?

The method depends on the equation, but each one ends with the arbitrary constants still visible, ready for the conditions. For first-order equations the main routes are:

• Direct integration when $\dfrac{dy}{dx} = f(x)$: the general solution is $y = \displaystyle\int f(x),dx + C$, built from antiderivatives.

• Separation of variables when the equation splits into $g(y),dy = f(x),dx$.

• Integrating factors for linear equations $y' + P(x)y = Q(x)$, set out step by step on the integrating factor method page.

• Characteristic equations for linear equations with constant coefficients, such as $y'' + y' - 6y = 0$, which gives $e^{2x}$ and $e^{-3x}$.

Sometimes the integral has no elementary antiderivative. Then the particular solution is written as an integral with a variable upper limit, which the fundamental theorem of calculus guarantees is a valid antiderivative. For $\dfrac{dy}{dx} = e^{-x^2}$ with $y(0) = 3$:

$$y(x) = 3 + \int_{0}^{x} e^{-t^2},dt$$

$$y(1) = 3 + \int_{0}^{1} e^{-t^2},dt \approx 3.7468$$

The integral of $e^{-t^2}$ is non-elementary, so this integral form, evaluated numerically, is the honest final answer.

Examples Of General And Particular Solutions

The six examples move from a one-step integration to a word problem, and every answer is checked by substitution.

Example 1

Find the general solution of $\dfrac{dy}{dx} = 6x^2 - 4$ and the particular solution through $(1, 3)$.

$$y = \int (6x^2 - 4),dx = 2x^3 - 4x + C$$

$$3 = 2(1)^3 - 4(1) + C$$

$$3 = -2 + C$$

$$C = 5$$

Check: $\dfrac{d}{dx}(2x^3 - 4x + 5) = 6x^2 - 4$, and the point gives $2 - 4 + 5 = 3$.

Final answer: general $y = 2x^3 - 4x + C$; particular $y = 2x^3 - 4x + 5$.

Example 2

Solve $\dfrac{dy}{dx} = y$ with $y(0) = 4$.

The tempting route separates and integrates, then attaches the constant at the end:

$$\frac{dy}{y} = dx$$

$$\ln|y| = x$$

$$y = e^x + C$$

$$4 = e^0 + C, \quad C = 3$$

That gives $y = e^x + 3$. Test it: $\dfrac{dy}{dx} = e^x$, but $y = e^x + 3$, and $e^x \neq e^x + 3$. The function fails the equation, so it cannot be a solution.

The break happened at the third line. The constant belongs to the integration step, before exponentiating:

$$\ln|y| = x + C$$

$$|y| = e^{x + C} = e^C e^x$$

$$y = Ae^x$$

where $A$ stands for $\pm e^C$ (and $A = 0$ covers the solution $y = 0$). Now apply the condition:

$$4 = Ae^0, \quad A = 4$$

Check: $\dfrac{d}{dx}(4e^x) = 4e^x = y$, and $y(0) = 4$.

Final answer: $y = 4e^x$.

Example 3

Solve $\dfrac{d^2y}{dx^2} = 12x$ with $y(0) = 1$ and $y'(0) = -2$.

Two integrations bring two constants:

$$\frac{dy}{dx} = 6x^2 + C_1$$

$$y = 2x^3 + C_1 x + C_2$$

$$y'(0) = C_1 = -2$$

$$y(0) = C_2 = 1$$

Check: $y = 2x^3 - 2x + 1$ gives $y'' = 12x$, $y(0) = 1$ and $y'(0) = -2$.

Final answer: $y = 2x^3 - 2x + 1$.

Example 4

Given the general solution $y = C_1 e^{2x} + C_2 e^{-3x}$ of $y'' + y' - 6y = 0$, find the particular solution with $y(0) = 0$ and $y'(0) = 5$.

$$y(0) = C_1 + C_2 = 0$$

$$y'(0) = 2C_1 - 3C_2 = 5$$

From the first equation, $C_2 = -C_1$. Substitute into the second:

$$2C_1 + 3C_1 = 5$$

$$C_1 = 1, \quad C_2 = -1$$

Check: $y = e^{2x} - e^{-3x}$ gives $y(0) = 0$ and $y'(0) = 2 + 3 = 5$, and the family was verified in the checking section above.

Final answer: $y = e^{2x} - e^{-3x}$.

Example 5

Find every solution of $\dfrac{dy}{dx} = 3y^{2/3}$ that passes through $(2, 0)$.

The general solution is $y = (x + C)^3$. Check: $\dfrac{dy}{dx} = 3(x + C)^2$ and $3y^{2/3} = 3(x + C)^2$. At $(2, 0)$:

$$0 = (2 + C)^3$$

$$C = -2$$

That gives the particular solution $y = (x - 2)^3$. The singular solution $y = 0$ also passes through $(2, 0)$, and it is not a member of the family.

Final answer: two solutions, $y = (x - 2)^3$ and $y = 0$.

Example 6

A cup of tea at $90^\circ\text{C}$ cools in a $20^\circ\text{C}$ room according to $\dfrac{dT}{dt} = -0.1(T - 20)$, with $t$ in minutes. Find $T(t)$, the temperature after 10 minutes, and when the tea reaches $50^\circ\text{C}$.

The general solution is $T = 20 + Ae^{-0.1t}$. Apply $T(0) = 90$:

$$90 = 20 + A, \quad A = 70$$

$$T(t) = 20 + 70e^{-0.1t}$$

$$T(10) = 20 + 70e^{-1} \approx 45.7516$$

For $50^\circ\text{C}$:

$$50 = 20 + 70e^{-0.1t}$$

$$e^{-0.1t} = \frac{3}{7}$$

$$t = 10\ln\frac{7}{3} \approx 8.4730$$

Check: $\dfrac{dT}{dt} = -7e^{-0.1t}$ and $-0.1(T - 20) = -0.1 \cdot 70e^{-0.1t} = -7e^{-0.1t}$.

Final answer: $T(t) = 20 + 70e^{-0.1t}$; about $45.75^\circ\text{C}$ after 10 minutes; $50^\circ\text{C}$ after about 8.47 minutes.

How Do Families Of Curves Make The Idea Visual?

The general solution is a family of curves filling the plane, and a particular solution is the single curve through a chosen point. A slope field draws the equation itself: at each point $(x, y)$ it shows a short segment with slope $\dfrac{dy}{dx}$, and every solution curve follows those segments. The direction fields page builds these by hand.

• Through almost every point, one curve. For well-behaved first-order equations, each point of the plane lies on exactly one member of the family, which is why one initial condition is enough.

• Second-order families are richer. Infinitely many curves of $y = C_1\cos x + C_2\sin x$ pass through the origin, one for each starting slope, which is why a second-order problem needs a value and a slope.

• Singular solutions sit outside. In the envelope picture, the singular curve touches the family without belonging to it.

Where Do General And Particular Solutions Show Up In Real Life?

They show up wherever a model is built once and then applied to a specific case. The general solution captures the mechanism; measured starting data turn it into a forecast.

• Motion. Constant gravity gives $s(t) = -4.9t^2 + C_1 t + C_2$; release height and speed fix the constants, as in velocity and acceleration problems.

• Cooling and heating. Newton's law of cooling gives a family of temperature curves; the starting temperature selects one, as in Example 6.

• Populations, radioactivity and interest. The family $y = Ae^{kt}$ of exponential growth and decay becomes a specific forecast once the starting amount $A$ is measured.

• Circuits and springs. Second-order equations need two pieces of data, such as initial charge and current, or initial position and velocity.

Where Do Students Trip Up On General And Particular Solutions?

Most errors here are not integration errors. They are bookkeeping errors about the constants.

Mistake 1: Attaching The Constant After Exponentiating

Where it slips in: Separable equations whose solution involves $\ln|y|$, such as $\dfrac{dy}{dx} = ky$.

Don't do this: Write $\ln|y| = kx$ and then $y = e^{kx} + C$.

The correct way: Add $C$ at the integration step, $\ln|y| = kx + C$, so exponentiating turns it into a multiplier: $y = Ae^{kx}$.

The rusher meets this one most often: the constant gets bolted on at the end out of habit from plain integration, and the answer looks finished. Substituting back takes ten seconds and exposes it every time.

Mistake 2: Stopping At The Value Of C

Where it slips in: The last line of an initial value problem.

Don't do this: Box "$C = 6$" as the answer.

The correct way: Write the full particular solution, $y = x^2 + 6$, because the question asks for a function.

Mistake 3: Too Few Conditions For The Order

Where it slips in: Second-order problems given only $y(0)$.

Don't do this: Solve for one constant and leave the other in place, then call the result "the particular solution".

The correct way: Count constants against conditions first. Two constants need two conditions, typically $y(x_0)$ and $y'(x_0)$; with only one, the honest answer is still a one-parameter family.

Mistake 4: Losing A Solution By Dividing

Where it slips in: Separating variables by dividing by $y$, $y^{2/3}$ or a similar factor.

Don't do this: Divide without asking whether the divisor can be zero.

The correct way: Before dividing, test whether setting the divisor to zero gives a solution. For $\dfrac{dy}{dx} = 3y^{2/3}$, $y = 0$ works, and it is the singular solution.

The student who memorised "general solution means all solutions" is the one this mistake catches, because the singular solution contradicts the slogan. In tutoring, the question that settles it is "what did you divide by, and could it be zero?", asked before any integration starts.

Practice Problems On General And Particular Solutions

1. Find the particular solution of $\dfrac{dy}{dx} = 4x^3 + 2$ with $y(0) = -1$.

2. Solve $\dfrac{dy}{dx} = -2y$ with $y(0) = 5$.

3. Solve $\dfrac{d^2y}{dx^2} = 6$ with $y(0) = 2$ and $y'(0) = 1$.

4. Form the differential equation whose general solution is $y = Ce^{3x}$.

5. The general solution of $y'' + 4y = 0$ is $y = C_1\cos 2x + C_2\sin 2x$. Find the particular solution with $y(0) = 3$ and $y'(0) = 4$.

6. Solve $\dfrac{dy}{dx} = \dfrac{x}{y}$ with $y(0) = 2$.

Answer to Question 1: $y = x^4 + 2x - 1$.

Answer to Question 2: $y = 5e^{-2x}$.

Answer to Question 3: $y = 3x^2 + x + 2$.

Answer to Question 4: $\dfrac{dy}{dx} = 3y$.

Answer to Question 5: $y = 3\cos 2x + 2\sin 2x$.

Answer to Question 6: $y = \sqrt{x^2 + 4}$ (the positive branch, since $y(0) = 2 > 0$).

Key Takeaways

• General and particular solutions differ only in their constants: the general one keeps them, the particular one fixes them.

• An $n$th-order equation has $n$ essential arbitrary constants, so it needs $n$ conditions to give one particular solution.

• Find the general solution first, substitute the conditions, then write the full function and check it by substitution.

• A singular solution satisfies the equation but comes from no choice of constants, often because a solving step divided by zero.

• In second-order linear work, "a particular solution" $y_p$ and "the particular solution of an initial value problem" are different objects.

What Should You Practise Next?

Work through problems 1 to 3 first; if the constant lands in the wrong place, return to Example 2, and if problem 5 stalls, reread the section on second-order particular solutions.

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