CBSE Sample Paper 2020-21 · Part B: Section IV · 3 marks

Q35: General solution of $x\,dy - (y + 2x^2)\,dx = 0$

Sample 2020-21Ch 9 · Differential Equations

The question

Find the general solution of the following differential equation:

$x\,dy - (y + 2x^2)\,dx = 0$

Exam answer

The given differential equation can be written as

\frac{dy}{dx} = \frac{y + 2x^2}{x} \Rightarrow \frac{dy}{dx} - \frac1x y = 2x

Why?

Move $(y + 2x^2)\,dx$ to the right and divide both sides by $x\,dx$ ⁠. Then split the fraction: $\frac{y + 2x^2}{x} = \frac{y}{x} + 2x$ ⁠, and bring $\frac{y}{x}$ to the left.

\begin{aligned} x\,dy &= (y + 2x^2)\,dx \\ \frac{dy}{dx} &= \frac{y + 2x^2}{x} \\ &= \frac{y}{x} + \frac{2x^2}{x} = \frac{y}{x} + 2x \\ \frac{dy}{dx} - \frac{y}{x} &= 2x \end{aligned}

This is of the form $\frac{dy}{dx} + Py = Q$ ⁠, with $P = -\frac1x$ ⁠, $Q = 2x$ ⁠.

Why?

It is linear in $y$ ⁠: $y$ and $\frac{dy}{dx}$ appear only to the first power and are not multiplied together. $P$ is the whole coefficient of $y$ ⁠, sign included.

Watch out: $P = -\frac1x$ ⁠, not $\frac1x$ ⁠.

\begin{aligned} \text{I.F.} &= e^{\int P\,dx} = e^{-\int\frac1x\,dx} \\ &= e^{-\log x} = \frac1x \end{aligned}

Why?

$\int\frac1x\,dx = \log x$ (taking $x > 0$ ⁠). Then use $-\log x = \log\frac1x$ and $e^{\log t} = t$ ⁠:

e^{-\log x} = e^{\log\frac1x} = \frac1x

The solution is

\begin{aligned} y \times \frac1x &= \int\left(2x \times \frac1x\right)dx \\ &= \int 2\,dx \\ \Rightarrow \frac{y}{x} &= 2x + C \end{aligned}

Why?

The solution of a linear equation is

y \times \text{I.F.} = \int Q \times \text{I.F.}\,dx + C

Here $2x \times \frac1x = 2$ ⁠, and $\int 2\,dx = 2x$ ⁠. Multiplying both sides by $x$ gives $y = 2x^2 + Cx$ ⁠.

\begin{aligned} \frac{y}{x} &= 2x + C \\ y &= x(2x + C) \quad (\times x) \\ &= 2x^2 + Cx \end{aligned}

$\therefore\ y = 2x^2 + Cx$ is the general solution.

Where this comes from

Chapter 9, Differential Equations: Linear differential equation: integrating factor.

Question type: Linear equations

How to spot it: click to see

Type 7. Linear equations: the integrating factor

Spot it: $\frac{dy}{dx} + Py = Q$ ⁠, with $P$ and $Q$ in $x$ only, and $y$ only to power 1.

Here: Divide by $x\,dx$ ⁠: $\frac{dy}{dx} - \frac1x y = 2x$ ⁠, with $y$ to power 1. So $P = -\frac1x$ ⁠, $\text{I.F.} = \frac1x$ ⁠, and $\frac{y}{x} = \int 2\,dx$ ⁠.

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Differential Equations in past papers: question 31 of 31 · see them all

Q35: General solution of x dy−(y+2x2) dx=0x\,dy - (y + 2x^2)\,dx = 0xdy−(y+2x2)dx=0

Q35: General solution of $x\,dy - (y + 2x^2)\,dx = 0$