Board 2023 Class 12 Maths Q15 Solution: General solution of x dy − (1 + x²) dx = dx

The question

The general solution of the differential equation $x\,dy - (1 + x^2)\,dx = dx$ is :

(a) $y = 2x + \frac{x^3}{3} + C$

(b) $y = 2\log x + \frac{x^3}{3} + C$

(c) $y = \frac{x^2}{2} + C$

(d) $y = 2\log x + \frac{x^2}{2} + C$

Exam answer

\begin{aligned} &x\,dy - (1 + x^2)\,dx = dx \\ \Rightarrow\ &x\,dy = dx + (1 + x^2)\,dx \\ \Rightarrow\ &x\,dy = (2 + x^2)\,dx \\ \Rightarrow\ &dy = \left(\frac{2}{x} + x\right)dx \end{aligned}

Why?

Collect all the $dx$ terms on one side: $1 + (1 + x^2) = 2 + x^2$ ⁠.

(Add $(1 + x^2)\,dx$ to both sides. Then $dx = 1\,dx$ ⁠, so the brackets add as $1 + 1 + x^2$ ⁠.)

Then divide by $x$ to separate the variables, with $y$ on the left and $x$ on the right:

\begin{aligned} \frac{2 + x^2}{x} &= \frac{2}{x} + \frac{x^2}{x} \\ &= \frac{2}{x} + x \end{aligned}

Integrating both sides:

\int dy = \int\left(\frac{2}{x} + x\right)dx

y = 2\log x + \frac{x^2}{2} + C

Why?

The standard integrals are

\begin{aligned} \int\frac{1}{x}\,dx &= \log|x| \\ \int x\,dx &= \frac{x^2}{2} \end{aligned}

Integrate each part on its own:

\begin{aligned} \int dy &= y \\ \int \frac{2}{x}\,dx &= 2\int\frac{1}{x}\,dx \\ &= 2\log|x| \\ \int x\,dx &= \frac{x^2}{2} \end{aligned}

The options write $\log x$ ⁠, which is the same as $\log|x|$ for $x > 0$ ⁠. One constant $C$ covers both sides.

Watch out: $\int x\,dx$ is $\frac{x^2}{2}$ ⁠, not $\frac{x^3}{3}$ ⁠. That slip gives option (b).

Answer: (d) $y = 2\log x + \dfrac{x^2}{2} + C$

Where this comes from

Chapter 9, Differential Equations: Variable separable differential equations.

Question type: Variables separable

How to spot it: click to see

Type 3. Variables separable: $y$ on one side, $x$ on the other

Spot it: $\frac{dy}{dx}$ equals a part in $x$ times a part in $y$ ⁠, like $(1 + x^2)(1 + y^2)$ ⁠.

Here: Collect the $dx$ terms: $x\,dy = (2 + x^2)\,dx$ ⁠. Divide by $x$ to get $dy = \left(\frac{2}{x} + x\right)dx$ ⁠, and integrate both sides.

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Q15: General solution of x dy−(1+x2) dx=dxx\,dy - (1 + x^2)\,dx = dxxdy−(1+x2)dx=dx