Solve: $x\frac{dy}{dx} + y = y^2\log x$.

Solve: x(dy)/(dx) + y = y²log x.

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x\frac{dy}{dx} + y = y^2\log x

Bihar BsebBihar Board Intermediate 2025Subjective· 5mImportance★★★★★

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Concept understanding — Nonlinear Differential Equations

Nonlinear Differential Equations

The idea

A differential equation tells you how a quantity changes. It is called linear when the unknown function and its derivatives appear only to the first power and are never multiplied together — that is the clean, well-behaved case NCERT's methods (variable-separable, homogeneous, linear first-order) are built around. A nonlinear differential equation is simply any equation that breaks this rule.

Note

Recognising whether an equation is linear or nonlinear is the classification step that decides which solving method can even be attempted — the integrating-factor method, for instance, works only for a linear first-order equation.

The precise statement

A differential equation is linear if it can be written as

a_n(x)\frac{d^ny}{dx^n}+\dots+a_1(x)\frac{dy}{dx}+a_0(x)\,y=f(x),

where the coefficients $a_i(x)$ and $f(x)$ depend only on the independent variable $x$ (or are constants), and $y$ together with its derivatives appears only to the first power and is never multiplied together.

A nonlinear differential equation violates at least one of these conditions:

a derivative or $y$ raised to a power other than $1$ , e.g. $y^2$ or $\left(\dfrac{dy}{dx}\right)^{2}$ ;
a product of $y$ with one of its derivatives, e.g. $y\dfrac{dy}{dx}$ ;
$y$ appearing inside a nonlinear function, e.g. $\sin y$ , $e^{y}$ , $\log y$ .

Examples

Linear (for contrast):

\frac{dy}{dx}+2y=e^{-x}.

Nonlinear:

\frac{dy}{dx}=y^{2}\quad(y \text{ is squared}),

x\left(\frac{dy}{dx}\right)^{2}+y=0\quad(\text{the slope is squared}),

\frac{dy}{dx}=y(1-y)\quad(\text{a product of } y \text{ and } 1-y).

Watch out

Do not judge by how complicated an equation looks. The equation $\dfrac{dy}{dx}=x^{3}y$ is linear (the coefficient of $y$ is just a function of $x$ ), while the simpler-looking $\dfrac{dy}{dx}=y^{2}$ is nonlinear.

Why the distinction matters …

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This is a Bernoulli equation; the substitution $v=1/y$ linearises it, and the integrating factor $1/x$ gives $\tfrac1y=1+\log x+Cx$ .

Start from $x\dfrac{dy}{dx} + y = y^2\log x.$ Divide by $x$ :

$\dfrac{dy}{dx} + \dfrac{y}{x} = \dfrac{y^2\log x}{x}.$

Bernoulli form ( $n=2$ ). Divide by $y^2$ :

$y^{-2}\dfrac{dy}{dx} + \dfrac{1}{x}y^{-1} = \dfrac{\log x}{x}.$

Let $v = y^{-1}$ , so $\dfrac{dv}{dx} = -y^{-2}\dfrac{dy}{dx}$ . Substituting:

$-\dfrac{dv}{dx} + \dfrac{1}{x}v = \dfrac{\log x}{x} \ \Rightarrow\ \dfrac{dv}{dx} - \dfrac{1}{x}v = -\dfrac{\log x}{x}.$

Integrating factor: $\mu = e^{-\int \frac{1}{x}dx} = e^{-\log x} = \dfrac{1}{x}.$

Then $\dfrac{d}{dx}\!\left(\dfrac{v}{x}\right) = \dfrac{1}{x}\cdot\left(-\dfrac{\log x}{x}\right) = -\dfrac{\log x}{x^2}.$

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Q.Solve: xdydx+y=y2log⁡xx\frac{dy}{dx} + y = y^2\log xxdxdy​+y=y2logx.

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Q.Solve: $x\frac{dy}{dx} + y = y^2\log x$ .

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