Differential equations with variables separable - Practice Questions & MCQ
Quick Facts
-
75 Questions around this concept.
Solve by difficulty
EASY If
and y(0)=1,then is equal to :
If is equal to
If $\frac{d y}{d x}=1+x+y+x y$, then
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Particular solution of D.E $e^{\frac{d y}{d x}}=x+2$ when $\mathrm{x}=-1, \mathrm{y}=2$
Solve $(x+y)^2 \frac{d y}{d x}=4$
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?
Which of the following does not represent variable separable form ?
For $a>0,{ }^{t \epsilon}\left[0, \frac{\pi}{2}\right]$, let $x=\sqrt{a^{\sin ^{-1} t}}$ and $y=\sqrt{a^{\cos ^{-1} t}}$. Then, $1+\left[\frac{d y}{d x}\right]^2$ equals to :
The general solution of differential equation $\left(e^{x}+1\right) y d y=(y+1) e^{x} d x$ is
Concepts Covered - 2
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}
$
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).
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