Verify: $\int \dfrac{2x+3}{x^2+3x}\,dx = \log|x^2+3x| + C$

Verify: ∫ (2x+3)/(x²+3x)dx = log|x²+3x| + C

NCERT Exemplar · Q2

\int \dfrac{2x+3}{x^2+3x}\,dx = \log|x^2+3x| + C

Assam AhsecShort· 3mImportance★★★★★est

82% · 304/373 Questions

✓ Free question

Concept understanding — Verification of Solution

Verifying a Solution of a Differential Equation

A function $y = \phi(x)$ is called a solution of a differential equation if, when you substitute it and its derivatives into the equation, the two sides become equal for every $x$ in the domain. Verification is the act of carrying out that substitution and checking that it holds as an identity.

The useful point: you do not have to solve the equation to verify a candidate. You are only checking a function that is already handed to you — which is exactly how many exam questions are phrased: "Show that $\dots$ is a solution of $\dots$ ."

The steps

From the given $y = \phi(x)$ , compute exactly the derivatives that appear in the equation.
Substitute $y$ and those derivatives into the left-hand side.
Simplify and check that it equals the right-hand side for all $x$ (an identity, not just at one point).

Example 1

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

Here $y' = -3e^{-3x}$ and $y'' = 9e^{-3x}$ . Substituting:

9e^{-3x} + (-3e^{-3x}) - 6e^{-3x} = (9 - 3 - 6)e^{-3x} = 0.

The left side is $0$ for every $x$ , so $y = e^{-3x}$ is a solution.

Example 2 (a solution with constants)

Verify that $y = a\cos x + b\sin x$ satisfies $\dfrac{d^2y}{dx^2} + y = 0$ for any constants $a, b$ .

Since $y'' = -a\cos x - b\sin x = -y$ , we get $y'' + y = 0$ . It holds for all $a, b$ , so this two-constant family is a solution.

Note

Verification links your answer back to the definition of a solution: a function is a solution not because of how you found it, but because it makes the differential equation true. If the substitution does not reduce to an identity, the function is simply not a solution.

Verifying that a given function solves a differential equation is explicitly listed as an exercise type in the NCERT Class 12 Mathematics textbook's Differential Equations chapter, and "verify the solution of differential equation examples" is a common CBSE and JEE Main search. This is often the easiest full-mark question in the chapter once the substitution steps are practiced a few times.

True. The numerator is the derivative of the denominator, so $\displaystyle\int\frac{2x+3}{x^2+3x}\,dx=\log|x^2+3x|+C$ ; differentiating the right side confirms it.

Whenever an integrand has the shape $\dfrac{u'(x)}{u(x)}$ , its antiderivative is $\log|u(x)|$ . Verifying is even simpler: differentiate the claimed answer and check you land back on the integrand.

Spot the pattern

Here $u=x^2+3x$ , and $u'=2x+3$ — which is exactly the numerator. So the integrand is $\dfrac{u'}{u}$ , and the natural antiderivative is $\log|u|=\log|x^2+3x|$ .

Differentiate to confirm

Let $F(x)=\log|x^2+3x|+C$ . By the chain rule,

F'(x)=\frac{1}{x^2+3x}\cdot\frac{d}{dx}(x^2+3x)=\frac{2x+3}{x^2+3x}.

This is the original integrand, and both are defined for $x\neq 0,-3$ , so the domains match.

Note

In calculus $\log$ denotes the natural logarithm $\log$ ; the derivative of $\log|u|$ is $u'/u$ , which is what makes the check work.

✓Final answer

\dfrac{d}{dx}\log|x^2+3x|=\dfrac{2x+3}{x^2+3x}

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AHSEC Higher Secondary (HS) Final Examination 2024Set ANNUAL4 marks
$\left(\dfrac{d^2y}{dx^2}\right)^5 + \left(\dfrac{dy}{dx}\right)^2 + \cos\left(\dfrac{dy}{dx}\right) + 1 = 0$
›Reveal solutionSolution

The ODE has order 2 but no defined degree (it isn't a polynomial in derivatives); $y=\sqrt{1+x^2}$ checks out as a genuine solution of the given first-order ODE. The OR part solves a separable ODE with an initial condition.

Order and degree: In $\left(\dfrac{d^2y}{dx^2}\right)^5+\left(\dfrac{dy}{dx}\right)^2+\cos\left(\dfrac{dy}{dx}\right)+1=0$ , the highest-order derivative is $d^2y/dx^2$ , so the order is 2. Degree is defined only when the equation is a polynomial in the derivatives; here $\cos(dy/dx)$ is a transcendental function of $dy/dx$ , not a polynomial term, so the degree is not defined.

Verification: $y=\sqrt{1+x^2}\Rightarrow \dfrac{dy}{dx}=\dfrac{x}{\sqrt{1+x^2}}$ . The RHS of the given equation is

$\frac{xy}{1+x^2}=\frac{x\sqrt{1+x^2}}{1+x^2}=\frac{x}{\sqrt{1+x^2}}=\frac{dy}{dx}.$

Since LHS $=$ RHS, $y=\sqrt{1+x^2}$ is indeed a solution.

OR: $x(x^2-1)\dfrac{dy}{dx}=1\Rightarrow dy=\dfrac{dx}{x(x-1)(x+1)}$ . Partial fractions:

$\frac{1}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1},$

giving (by substitution) $A=-1,\ B=\tfrac12,\ C=\tfrac12$ . So

$y=-\ln|x|+\frac12\ln|x-1|+\frac12\ln|x+1|+K=-\ln|x|+\frac12\ln|x^2-1|+K.$

Using $y=0$ at $x=2$ : $0=-\ln2+\tfrac12\ln3+K\Rightarrow K=\ln2-\tfrac12\ln3$ . So the particular solution is

$y=-\ln x+\frac12\ln(x^2-1)+\ln2-\frac12\ln3=\ln\frac{2}{x}+\frac12\ln\frac{x^2-1}{3}\quad(x>2\text{ region}).$

✓Final answer
$=2$
AHSEC Higher Secondary (HS) Final Examination 2019Set ANNUAL4 marks
$y = 3\cos(\log x) + 4\sin(\log x)$
›Reveal solutionSolution

Differentiate twice using the chain rule, multiply by $x$ and $x^2$ appropriately, and the terms cancel to give $0$ .

$y = 3\cos(\log x) + 4\sin(\log x)$ .

First derivative (chain rule, $\frac{d}{dx}\log x = 1/x$ ):

$y' = -3\sin(\log x)\cdot\frac1x + 4\cos(\log x)\cdot\frac1x = \frac1x\left[4\cos(\log x) - 3\sin(\log x)\right].$

So $xy' = 4\cos(\log x) - 3\sin(\log x)$ .

Differentiate $xy'$ again with respect to $x$ (product rule on the LHS, chain rule on the RHS):

$y' + xy'' = -4\sin(\log x)\cdot\frac1x - 3\cos(\log x)\cdot\frac1x = -\frac1x\left[3\cos(\log x)+4\sin(\log x)\right] = -\frac{y}{x}.$

Multiply both sides by $x$ :

$xy' + x^2y'' = -y.$

Rearranging:

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

This is exactly the required relation, so it is proved.

✓Final answer
$x^2y''+xy'+y=0$

←Previous questionQ1 — Verify: ∫ (2x-1)/(2x+3)dx = x - log|(2x+3)²| + C →Next questionQ3 — Evaluate: ∫ (x²+2)/(x+1)dx

Q.Verify: $\int \dfrac{2x+3}{x^2+3x}\,dx = \log|x^2+3x| + C$

Concept understanding — Verification of Solution

Verifying a Solution of a Differential Equation

The steps

Example 1

Example 2 (a solution with constants)

Spot the pattern

Differentiate to confirm

Method: Verifying $\int \frac{f'(x)}{f(x)}\,dx=\log|f(x)|+C$ by differentiation

Steps

Common Mistakes

Q.Verify: ∫2x+3x2+3x dx=log⁡∣x2+3x∣+C\int \dfrac{2x+3}{x^2+3x}\,dx = \log|x^2+3x| + C∫x2+3x2x+3​dx=log∣x2+3x∣+C

Spot the pattern

Differentiate to confirm

🎓Unlock everything free for 14 days

Q.Verify: $\int \dfrac{2x+3}{x^2+3x}\,dx = \log|x^2+3x| + C$

Unlock everything free for 14 days