General and Particular Solutions

What Does It Mean to Solve a Differential Equation?

For an ordinary equation like $x^2 + 1 = 0$ or $\sin^2 x - \cos x = 0$, a solution is a number that satisfies it. For a differential equation like

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

a solution is a function $y = \phi(x)$: substituting $\phi$ and its derivatives for $y$ makes L.H.S. $=$ R.H.S. for every $x$. The graph $y = \phi(x)$ is called the solution curve (or integral curve) of the equation.

General versus particular

Consider $y = \phi(x) = a\sin(x + b)$, with $a, b \in \mathbb{R}$. Substituting into $y'' + y = 0$ gives L.H.S. $=$ R.H.S. — a solution for every choice of $a$ and $b$. This two-parameter family is the general solution (also called the primitive).

Now fix the parameters — say $a = 2$, $b = \frac{\pi}{4}$ — to get $y = \phi_1(x) = 2\sin\left(x + \frac{\pi}{4}\right)$: still a solution, but with no arbitrary constants. That is a particular solution.

Definitions: The solution containing arbitrary constants is the general solution. A solution free from arbitrary constants — obtained by giving the constants particular values — is a particular solution.

Counting the constants

The general solution of an $n$th order differential equation contains exactly $n$ arbitrary constants — one integration per order. So a fourth-order equation's general solution carries $4$ constants, while any particular solution carries $0$ (that is what "particular" means).

Key Point: One differential equation ↔ one family of curves. Extra information — "the curve passes through $(1, 2)$" — pins down the constants and selects one member of the family. This is exactly how the solving methods of the next three sections will be used in exams: general solution first, then apply the condition.

[JEE Tip] Constant-counting is a free-marks MCQ: general solution of order $n$ → $n$ constants; particular solution of any order → $0$ constants. No computation, just the definitions.

Verifying a Solution — the Substitution Recipe

Verification questions hand you a function and an equation and ask you to check the fit. No solving techniques needed — just careful differentiation.

The recipe:

1. Differentiate the given function as many times as the order of the equation demands.
2. Substitute the function and its derivatives into the L.H.S.
3. Simplify and confirm L.H.S. $=$ R.H.S. identically (for all $x$ in the domain, not just at one point).

Worked template

Verify that $y = e^{-3x}$ solves $\dfrac{d^2y}{dx^2} + \dfrac{dy}{dx} - 6y = 0$:

1. $\frac{dy}{dx} = -3e^{-3x}$, and differentiating again, $\frac{d^2y}{dx^2} = 9e^{-3x}$.
2. Substitute: L.H.S. $= 9e^{-3x} + \left(-3e^{-3x}\right) - 6e^{-3x} = 9e^{-3x} - 9e^{-3x} = 0 =$ R.H.S. ✓

Implicit solutions

Solutions sometimes arrive as implicit relations, like $xy = \log y + C$. Differentiate the relation implicitly:

$$y + xy' = \frac{y'}{y} \;\Rightarrow\; y'\left(x - \frac1y\right) = -y \;\Rightarrow\; y' = \frac{y^2}{1 - xy}$$

which is exactly the differential equation to be verified. Implicit verification = implicit differentiation + algebra to isolate $y'$.

Key Point: Verification is differentiation running forward — no integrals anywhere. If the check leaves a leftover term, either the differentiation slipped or the function genuinely isn't a solution; both are findable by redoing step 1 slowly.

[JEE Tip] When a verification involves square roots (like $y = x\sin x$ against $xy' = y + x\sqrt{x^2 - y^2}$), the identity may hold only on a stated domain (here where $\cos x \geq 0$) — quote the domain restriction; both boards and JEE include it in the full-credit answer.

Solved Examples

Example 1: Second-order verification

Verify that $y = e^{-3x}$ is a solution of $\dfrac{d^2y}{dx^2} + \dfrac{dy}{dx} - 6y = 0$.

Solution:

1. Differentiate twice: $y' = -3e^{-3x}$; $y'' = 9e^{-3x}$.
2. Substitute: L.H.S. $= 9e^{-3x} - 3e^{-3x} - 6e^{-3x} = 0 =$ R.H.S.

Final Answer: Verified — $y = e^{-3x}$ is a solution.

Example 2: A two-parameter family

Verify that $y = a\cos x + b\sin x$ (with $a, b \in \mathbb{R}$) solves $\dfrac{d^2y}{dx^2} + y = 0$.

Solution:

1. Differentiate twice: $y' = -a\sin x + b\cos x$; $y'' = -a\cos x - b\sin x$.
2. Substitute: L.H.S. $= (-a\cos x - b\sin x) + (a\cos x + b\sin x) = 0$.

Final Answer: Verified for every $a$ and $b$ — and with two arbitrary constants matching the order $2$, this is the general solution.

Example 3: A root function

Verify that $y = \sqrt{1 + x^2}$ is a solution of $y' = \dfrac{xy}{1 + x^2}$.

Solution:

1. Differentiate: $y' = \frac{2x}{2\sqrt{1 + x^2}} = \frac{x}{\sqrt{1+x^2}}$.
2. Compute the R.H.S.: $\frac{xy}{1+x^2} = \frac{x\sqrt{1+x^2}}{1+x^2} = \frac{x}{\sqrt{1+x^2}}$.
3. Compare: L.H.S. $=$ R.H.S. ✓

Final Answer: Verified.

Takeaway: When both sides are computed independently and meet in the middle, the verification is airtight — rearranging one side into the other risks circular algebra.

Example 4: An implicit solution

Verify that $xy = \log y + C$ is a solution of $y' = \dfrac{y^2}{1 - xy}$ (where $xy \neq 1$).

Solution:

1. Differentiate implicitly: $\frac{d}{dx}(xy) = \frac{d}{dx}(\log y + C)$ gives $y + xy' = \frac{y'}{y}$.
2. Collect $y'$: $y'\left(\frac1y - x\right) = y$, so $y'\cdot\frac{1 - xy}{y} = y$.
3. Isolate: $y' = \frac{y^2}{1 - xy}$ — exactly the given equation.

Final Answer: Verified.

Example 5: A domain-sensitive verification

Verify that $y = x\sin x$ is a solution of $xy' = y + x\sqrt{x^2 - y^2}$ (for $x \neq 0$ and $x > y$ or $x < -y$).

Solution:

1. Differentiate: $y' = \sin x + x\cos x$, so $xy' = x\sin x + x^2\cos x = y + x^2\cos x$.
2. Compute the root: $\sqrt{x^2 - y^2} = \sqrt{x^2 - x^2\sin^2x} = \sqrt{x^2\cos^2x} = |x\cos x|$.
3. Compare: on the stated domain (where $x\cos x \geq 0$), $x\sqrt{x^2 - y^2} = x^2\cos x$, so R.H.S. $= y + x^2\cos x =$ L.H.S. ✓

Final Answer: Verified on the stated domain.

Takeaway: $\sqrt{u^2} = |u|$, not $u$ — the domain restriction in the problem statement is exactly what turns the modulus into a plain factor. Quote it.

Example 6: From general to particular

The general solution of $\dfrac{dy}{dx} = 2x$ is $y = x^2 + C$. Find the particular solution whose curve passes through $(1, 2)$.

Solution:

1. Apply the condition: substitute $x = 1$, $y = 2$: $2 = 1 + C$.
2. Solve for the constant: $C = 1$.
3. Write the particular solution: $y = x^2 + 1$.

Final Answer: $y = x^2 + 1$ — one curve selected from the one-parameter family.

Takeaway: This two-step move — solve generally, then impose the condition — is the closing move of nearly every solving problem in the rest of the chapter.