CBSE Sample Paper 2020-21 · Part A: Section I · 1 mark

Q9: Arbitrary constants in a particular solution. Or: $n$ for a homogeneous equation

Sample 2020-21Ch 9 · Differential Equations

The question

(a) How many arbitrary constants are there in the particular solution of the differential equation $\frac{dy}{dx} = -4xy^2$ ; $y(0) = 1$

(b) For what value of n is the following a homogeneous differential equation:

\frac{dy}{dx} = \frac{x^3 - y^n}{x^2y + xy^2}

Exam answer

(a)

Separating the variables and integrating,

\begin{aligned} \int\frac{dy}{y^2} &= -4\int x\,dx \\ -\frac{1}{y} &= -2x^2 + C \end{aligned}

Why?

Divide both sides by $y^2$ and multiply by $dx$ ⁠, so that each side has one variable only. Then integrate each side:

\begin{aligned} \frac{dy}{dx} &= -4xy^2 \\ \frac{dy}{y^2} &= -4x\,dx \end{aligned}

Left side, with $\int y^n\,dy = \frac{y^{n+1}}{n+1}$ and $n = -2$ ⁠:

\begin{aligned} \int y^{-2}\,dy &= \frac{y^{-2+1}}{-2+1} \\ &= \frac{y^{-1}}{-1} = -\frac{1}{y} \end{aligned}

Right side:

\begin{aligned} -4\int x\,dx &= -4 \cdot \frac{x^2}{2} \\ &= -2x^2 \end{aligned}

One constant $C$ appears, because the equation is of order 1.

Putting $x = 0$ ⁠, $y = 1$ ⁠: $-1 = C$ ⁠. So

y = \frac{1}{2x^2 + 1}

Why?

Put $x = 0$ ⁠, $y = 1$ into $-\frac{1}{y} = -2x^2 + C$ ⁠:

\begin{aligned} -\frac{1}{1} &= -2(0)^2 + C \\ -1 &= 0 + C \\ C &= -1 \end{aligned}

With $C = -1$ ⁠, $-\frac{1}{y} = -2x^2 - 1$ ⁠, so $\frac{1}{y} = 2x^2 + 1$ ⁠. Check: at $x = 0$ ⁠, $y = \frac{1}{1} = 1$ ⁠.

\begin{aligned} -\frac{1}{y} &= -2x^2 - 1 \\ \frac{1}{y} &= 2x^2 + 1 \quad (\times -1) \\ y &= \frac{1}{2x^2 + 1} \quad (\text{flip}) \end{aligned}

A particular solution has no arbitrary constant: the constant of the general solution is fixed by $y(0) = 1$ ⁠.

Why?

The general solution contains the arbitrary constant $C$ ⁠. A particular solution is the one curve picked out by the condition, so $C$ has a fixed value.

Watch out: the answer is not 1. One is the number of constants in the general solution.

Number of arbitrary constants $= 0$ ⁠.

(b)

The denominator $x^2y + xy^2$ is homogeneous of degree 3, so the numerator $x^3 - y^n$ must also be of degree 3, that is $n = 3$ ⁠.

Why?

The equation $\frac{dy}{dx} = F(x, y)$ is homogeneous when $F(\lambda x, \lambda y) = \lambda^0 F(x, y)$ for every $\lambda$ ⁠.

Each term of the denominator has total degree 3: $x^2y$ has $2 + 1$ and $xy^2$ has $1 + 2$ ⁠. The numerator has the term $x^3$ ⁠, so $y^n$ must have degree 3 as well.

Then, with $F(x, y) = \frac{x^3 - y^3}{x^2y + xy^2}$ ⁠,

F(\lambda x, \lambda y) = \frac{\lambda^3(x^3 - y^3)}{\lambda^3(x^2y + xy^2)}

Why?

Replacing $x$ by $\lambda x$ and $y$ by $\lambda y$ turns every term of degree 3 into $\lambda^3$ times that term. So $\lambda^3$ comes out of the top and of the bottom, and cancels.

\begin{aligned} \text{Top: }&(\lambda x)^3 - (\lambda y)^3 \\ &= \lambda^3x^3 - \lambda^3y^3 \\ &= \lambda^3(x^3 - y^3) \\[4pt] \text{Bottom: }&(\lambda x)^2(\lambda y) \\ &\quad + (\lambda x)(\lambda y)^2 \\ &= \lambda^3x^2y + \lambda^3xy^2 \\ &= \lambda^3(x^2y + xy^2) \end{aligned}

Cancel $\lambda^3$ from top and bottom: what is left is $F(x, y)$ ⁠, and $\lambda^0 = 1$ ⁠.

Watch out: with $n = 2$ ⁠, the top becomes $\lambda^3x^3 - \lambda^2y^2$ ⁠, and $\lambda$ does not cancel.

F(\lambda x, \lambda y) = \lambda^0 F(x, y)

$\therefore\ n = 3$

Where this comes from

Chapter 9, Differential Equations: Particular solution; homogeneous equation.

Question type: (a) A particular solution; (b) Homogeneous equations

How to spot it: click to see

(a) Type 4. A particular solution: solve, then find $C$

Spot it: a condition such as " $y = 1$ when $x = 0$ ", or "the curve passes through a point".

Here: The condition $y(0) = 1$ fixes the one constant of $-\frac{1}{y} = -2x^2 + C$ ⁠: $C = -1$ ⁠, giving $y = \frac{1}{2x^2 + 1}$ ⁠. No constant is left, so the answer is 0.

(b) Type 6. Homogeneous equations: put $y = vx$

Spot it: every term has the same degree, like $x^2$ ⁠, $xy$ and $y^2$ ⁠, or you see $\frac{y}{x}$ inside sin, cos or log. You cannot separate $x$ and $y$ ⁠.

Here: The bottom $x^2y + xy^2$ has every term of degree 3, so the top $x^3 - y^n$ must be degree 3 too: $n = 3$ ⁠, and then $F(\lambda x, \lambda y) = F(x, y)$ ⁠.

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Differential Equations in past papers: question 29 of 31 · see them all

Q9: Arbitrary constants in a particular solution. Or: nnn for a homogeneous equation

Q9: Arbitrary constants in a particular solution. Or: $n$ for a homogeneous equation