(vii) The solution of $\frac{dy}{dx}=\left(\frac{y}{x}\right)^{1/3}$ is $y^{2/3}-x^{2/3}=c$. (State True or False.)

> [!TLDR] > The given differential equation is homogeneous and can be solved by the substitution $y = vx$. After separating variables and integrating, the general solution is indeed $y^{2/3} - x^{2/3} = c$, so the statement is **True**. The key here is recognising the structure of the equation. When you see $\frac{dy}{dx}$ expressed purely as a function of $\frac{y}{x}$, you’re looking at a **homogeneous differential equation**. The standard trick — substitute $y = vx$ — turns it into a separable equation, which you can then integrate directly. Let’s walk through it. 1. **Rewrite the equation** We have $$ \frac{dy}{dx} = \left(\frac{y}{x}\right)^{1/3}. $$ The right-hand side depends only on the ratio $y/x$, confirming homogeneity. 2. **Substitute $y = vx$** Then $y = vx$ implies $\frac{dy}{dx} = v + x\frac{dv}{dx}$ (by the product rule). The equation becomes $$ v + x\frac{dv}{dx} = v^{1/3}. $$ 3. **Separate variables** Subtract $v$ from both sides: $$ x\frac{dv}{dx} = v^{1/3} - v. $$ Factor the right-hand side: $$ v^{1/3} - v = v^{1/3}(1 - v^{2/3}). $$ So $$ x\frac{dv}{dx} = v^{1/3}(1 - v^{2/3}). $$ Now separate: $$ \frac{dv}{v^{1/3}(1 - v^{2/3})} = \frac{dx}{x}. $$ 4. **Integrate both sides** The left side looks messy, but a clever substitution cleans it up. Let $t = v^{2/3}$. Then $v = t^{3/2}$, so $dv = \frac{3}{2} t^{1/2}\, dt$. Also $v^{1/3} = t^{1/2}$. Substituting: $$ \frac{dv}{v^{1/3}(1 - v^{2/3})} = \frac{\frac{3}{2} t^{1/2}\, dt}{t^{1/2}(1 - t)} = \frac{3}{2} \cdot \frac{dt}{1 - t}. $$ The integral becomes $$ \frac{3}{2} \int \frac{dt}{1 - t} = \int \frac{dx}{x}. $$ Integrating: $$ -\frac{3}{2} \log|1 - t| = \log|x| + C. $$ > [!TIP] > The substitution $t = v^{2/3}$ is the natural choice because the denominator has $1 - v^{2/3}$. It turns the integral into a standard logarithmic form. 5.

(vii) The solution of (dy)/(dx)=((y)/(x))1/3 is y2/3-x2/3=c. (State True or False.)

NCERT Exemplar · Q51

\frac{dy}{dx}=\left(\frac{y}{x}\right)^{1/3}

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The given differential equation is homogeneous and can be solved by the substitution $y = vx$ . After separating variables and integrating, the general solution is indeed $y^{2/3} - x^{2/3} = c$ , so the statement is True.

The key here is recognising the structure of the equation. When you see $\frac{dy}{dx}$ expressed purely as a function of $\frac{y}{x}$ , you’re looking at a homogeneous differential equation. The standard trick — substitute $y = vx$ — turns it into a separable equation, which you can then integrate directly.

Let’s walk through it.

Rewrite the equation We have

\frac{dy}{dx} = \left(\frac{y}{x}\right)^{1/3}.

The right-hand side depends only on the ratio $y/x$ , confirming homogeneity.

Substitute $y = vx$ Then $y = vx$ implies $\frac{dy}{dx} = v + x\frac{dv}{dx}$ (by the product rule). The equation becomes

v + x\frac{dv}{dx} = v^{1/3}.

Separate variables Subtract $v$ from both sides:

x\frac{dv}{dx} = v^{1/3} - v.

Factor the right-hand side:

v^{1/3} - v = v^{1/3}(1 - v^{2/3}).

x\frac{dv}{dx} = v^{1/3}(1 - v^{2/3}).

Now separate:

\frac{dv}{v^{1/3}(1 - v^{2/3})} = \frac{dx}{x}.

Integrate both sides The left side looks messy, but a clever substitution cleans it up. Let $t = v^{2/3}$ . Then $v = t^{3/2}$ , so $dv = \frac{3}{2} t^{1/2}\, dt$ . Also $v^{1/3} = t^{1/2}$ . Substituting:

\frac{dv}{v^{1/3}(1 - v^{2/3})} = \frac{\frac{3}{2} t^{1/2}\, dt}{t^{1/2}(1 - t)} = \frac{3}{2} \cdot \frac{dt}{1 - t}.

The integral becomes

\frac{3}{2} \int \frac{dt}{1 - t} = \int \frac{dx}{x}.

Integrating:

-\frac{3}{2} \log|1 - t| = \log|x| + C.

Tip

The substitution $t = v^{2/3}$ is the natural choice because the denominator has $1 - v^{2/3}$ . It turns the integral into a standard logarithmic form.

Back-substitute Recall $t = v^{2/3}$ and $v = y/x$ , so $t = (y/x)^{2/3} = y^{2/3} / x^{2/3}$ . Then $1 - t = 1 - \frac{y^{2/3}}{x^{2/3}} = \frac{x^{2/3} - y^{2/3}}{x^{2/3}}$ . The equation becomes

-\frac{3}{2} \log\left|\frac{x^{2/3} - y^{2/3}}{x^{2/3}}\right| = \log|x| + C.

Use logarithm properties:

-\frac{3}{2} \left[ \log|x^{2/3} - y^{2/3}| - \log|x^{2/3}| \right] = \log|x| + C.

But $\log|x^{2/3}| = \frac{2}{3}\log|x|$ , so …

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Q.(vii) The solution of $\frac{dy}{dx}=\left(\frac{y}{x}\right)^{1/3}$ is $y^{2/3}-x^{2/3}=c$ . (State True or False.)

Concept understanding — Verification of Solution

Verifying a Solution of a Differential Equation

The steps

Example 1

Example 2 (a solution with constants)

Method: Verifying a Proposed Solution by Implicit Differentiation

Steps

Q.(vii) The solution of dydx=(yx)1/3\frac{dy}{dx}=\left(\frac{y}{x}\right)^{1/3}dxdy​=(xy​)1/3 is y2/3−x2/3=cy^{2/3}-x^{2/3}=cy2/3−x2/3=c. (State True or False.)

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Q.(vii) The solution of $\frac{dy}{dx}=\left(\frac{y}{x}\right)^{1/3}$ is $y^{2/3}-x^{2/3}=c$ . (State True or False.)

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