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Differential equations with variables separable - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

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37 Questions around this concept.

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If $\left ( 2+\sin x \right )\: \frac{dy}{dx}\: +\left ( y+1 \right )\cos x= 0$

and y(0)=1,then $y\left ( \frac{\pi }{2} \right )$ is equal to :

$-\frac{2}{3}$

$-\frac{1}{3}$

$\frac{4}{3}$

$\frac{1}{3}$

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If $\dpi{100} \frac{dy}{dx}=y+3> 0\; and\; y(0)=2,then\; y(1n\: 2)$ is equal to

– 2

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Concepts Covered - 2

Differential equations with variables separable

$\\\mathrm{The \;differential\;of\;the\\;form\;\frac{dy}{dx}=f(x)g(y)\;where\;f(x)\;is\;function\;of\;x}\\\mathrm{and\;g(y)\;is\;a\;function\;of\;y,\;are\;said\;to\;be\;variable\;separrable\;form.}$

Rewrite the equation as

$\\\mathrm{\frac{dy}{g(y)}=f(x)dx\;\;\;\;\;\;\;\;[where\;g(y)\neq0]}\\\\\text{This\;process is separating the variables.}\\\mathrm{Now,\; integrate\;both\;side,\;we\;get}\\\\\mathrm{\int \frac{dy}{g(y)}=\int f(x)\;dx+c}$

By this, we get the solution of differential equation

Let’s see one illustration for better understanding

Solution of the differential equation $\\\mathrm{\frac{dy}{dx}=(e^x+1)(y^2+1)}$

Rewrite the differential equation as

$\\ \frac{\mathrm{dy}}{1+\mathrm{y}^{2}}=\left(\mathrm{e}^{\mathrm{x}}+1\right) \mathrm{d} \mathrm{x}\\\\\text{Integrating both sides, we get }\\\\ \int\frac{\mathrm{dy}}{1+\mathrm{y}^{2}}=\int \left(\mathrm{e}^{\mathrm{x}}+1\right) \mathrm{d} \mathrm{x} \\\Rightarrow \tan ^{-1} y=e^{x}+x+c\\\Rightarrow y=\tan \left(e^{x}+x+c\right)$

Differential Equation Reducible to Variable Separable Form

A differential equation of the form $\\\mathrm{\frac{dy}{dx}=f(ax+by+c)}$ where a, b and c are constants, can be converted into an equation with variables separable by the substitution v = ax + by + c.

$\\\mathrm{\frac{dy}{dx}=f(ax+by+c)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(i)}\\\\\mathrm{v = ax + by + c.}\\\therefore \quad \frac{d v}{d x}=a+b \frac{d y}{d x} \text { or, } \frac{d y}{d x}=\frac{\frac{d v}{d x}-a}{b}\\\\\Rightarrow \;\;\frac{\frac{\mathrm{d} v}{\mathrm{dx}}-\mathrm{a}}{\mathrm{b}}=\mathrm{f}(\mathrm{v}) \Rightarrow \frac{\mathrm{d} \mathrm{v}}{\mathrm{dx}}=\mathrm{bf}(\mathrm{v})+\mathrm{a}\\\\\Rightarrow \;\;\;\frac{\mathrm{d} \mathrm{v}}{\mathrm{bf}(\mathrm{v})+\mathrm{a}}=\mathrm{dx}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots\mathrm{(ii)}\\\\\text{In the differential equation (ii), the variables x and v are separated.}\\\text{Integrating (ii), we get}\\\Rightarrow \;\;\;\;\int \frac{\mathrm{dv}}{\mathrm{bf}(\mathrm{v})+\mathrm{a}}=\int \mathrm{dx}+\mathrm{C}\\\\\Rightarrow \;\;\;\int \frac{\mathrm{d} \mathrm{v}}{\mathrm{bf}(\mathrm{v})+\mathrm{a}}=\mathrm{x}+\mathrm{C}, \text { where } \mathrm{v}=\mathrm{ax}+\mathrm{by}+\mathrm{c}$

This represents the general solution of the differential equation (i).

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Differential equations with variables separable

Differential Equation Reducible to Variable Separable Form

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Reference Books

Differential equations with variables separable

Mathematics for Joint Entrance Examination JEE (Advanced) : Calculus

Page No. : 10.4

Line : 45

Differential Equation Reducible to Variable Separable Form

Mathematics for Joint Entrance Examination JEE (Advanced) : Calculus

Page No. : 10.4

Line : 47

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