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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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  • 78 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 :

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

Given $y(0)=2000$ and $\frac{d y}{d x}=32000-20 y^2$, find the value of $\lim _{x \rightarrow \infty} y(x)$.

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If $\frac{d y}{d x}=1+x+y+x y$, then

A function $y=f(x)$ satisfies $x f^{\prime}(x)+2 f(x)=2 x \sec ^2 x \sqrt{f(x)}$ with $f(0)=1$. Then find the value of $f(2 \pi)$.

Particular solution of D.E $e^{\frac{d y}{d x}}=x+2$ when $\mathrm{x}=-1, \mathrm{y}=2$

Solve $(x+y)^2 \frac{d y}{d x}=4$

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The solution of D.E

$\left ( ye^{xy}+\frac{1}{y}e^{x/y} \right )xdy=\left (e^{x/y} -y^2e^{xy} \right )dx$  is

Which of the following can be solved using variable separable method?

 

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Which of the following does not represent variable separable form ?

Concepts Covered - 2

Differential equations with variables separable

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

Rewrite the equation as
$
\frac{d y}{g(y)}=f(x) d x \quad[\text { where } g(y) \neq 0]
$

This process is separating the variables.
Now, integrating both sides, we get
$
\int \frac{\mathrm{dy}}{\mathrm{~g}(\mathrm{y})}=\int \mathrm{f}(\mathrm{x}) \mathrm{dx}+\mathrm{c}
$

By this, we get the solution of the differential equation

Let’s see one illustration for a better understanding

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

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

 

Differential Equation Reducible to Variable Separable Form

A differential equation of the form $\frac{d y}{d x}=f(a x+b y+c)$ where $\mathrm{a}, \mathrm{b}$ and c are constants, can be converted into an equation with variables separable by the substitution $\mathrm{v}=\mathrm{ax}+\mathrm{by}+\mathrm{c}$.
$
\begin{aligned}
& \frac{d y}{d x}=f(a x+b y+c) \\
& \mathrm{v}=\mathrm{ax}+\mathrm{by}+\mathrm{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} v}{\mathrm{dx}}=\mathrm{bf}(\mathrm{v})+\mathrm{a} \\
& \Rightarrow \frac{\mathrm{dv}}{\mathrm{bf}(\mathrm{v})+\mathrm{a}}=\mathrm{dx}
\end{aligned}
$

In the differential equation (ii), the variables x and v are separated.
Integrating (ii), we get
$
\begin{aligned}
& \Rightarrow \quad \int \frac{\mathrm{dv}}{\mathrm{bf}(\mathrm{v})+\mathrm{a}}=\int \mathrm{dx}+\mathrm{C} \\
& \Rightarrow \quad \int \frac{\mathrm{dv}}{\mathrm{bf}(\mathrm{v})+\mathrm{a}}=\mathrm{x}+\mathrm{C}, \text { where } \mathrm{v}=\mathrm{ax}+\mathrm{by}+\mathrm{c}
\end{aligned}
$

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

Study it with Videos

Differential equations with variables separable
Differential Equation Reducible to Variable Separable Form

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Books

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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