Question

If the given equation is of the form $\dfrac{{dy}}{{dx}} - y\cot x = {\text{cosec }}x$, then its solution is $y \cdot {\text{cosec }}x = \cot x + c$. If true enter 1 else 0.

Answer 1

Hint : In this question, we need to determine the solution of the differential equation $\dfrac{{dy}}{{dx}} - y\cot x = {\text{cosec }}x$ and compare the result with the given solution $y \cdot {\text{cosec }}x = \cot x + c$. If the calculated solution is equivalent to the given solution then, the answer is 1 otherwise 0. For this, we will first find the integrating factor and then find the general solution equation by multiplying the integrating factor.

Complete step-by-step answer :
Given the differential equation is $\dfrac{{dy}}{{dx}} - y\cot x = {\text{cosec }}x$
This equation can be written as $\dfrac{{dy}}{{dx}} - \cot x.y = {\text{cosec }}x - - (i)$
We know the standard linear differential equation is $\dfrac{{dy}}{{dx}} + P(x)y = Q(x)$
So by comparing equation (i) by standard linear differential equation we can say
$P(x) = - \cot x$
$Q(x) = {\text{cosec }}x$
Now we find the Integrating factor of the differential equation which is given by the formula
$IF = {e^{\int {Pdx} }} - - (ii)$
Where$P(x) = - \cot x$, now substitute this value in equation (ii), we get

$$IF = {e^{\int {Pdx} }} \\\ = {e^{ - \int {\cot x} }} \\\ = {e^{ - \ln \sin x}} \\\ = {e^{\ln {{\sin }^{ - 1}}x}} \\\ = {e^{\ln \left( {\dfrac{1}{{\sin x}}} \right)}} \\\ = \dfrac{1}{{\sin x}} \\\ = {\text{cosec }}x \\\$$

Hence we get the Integrating factor of the linear equation$= \cos ecx$
Now we know the general solution of the linear differential equation is
$y.IF = \int {\left( {Q(x).IF} \right)dx} + c - - (iii)$
As we already know the value of $Q(x) = {\text{cosec }}x$ and $IF = {\text{cosec }}x$, hence by substituting these values in equation (iii) we can write
$y \cdot {\text{cosec }}x = \int {\left( {{\text{cosec }}x \cdot {\text{cosec }}x} \right)dx} + c$
By solving this we get

$$y \cdot {\text{cosec }}x = \int {{\text{cose}}{{\text{c}}^2}xdx} + c \\\ y \cdot {\text{cosec}}x = - \cot x + c \\\$$

As we get the solution for the differential equation as $y \cdot {\text{cosec }}x = - \cot x + c$ which is not equivalent to the given solution of the differential solution $y \cdot {\text{cosec }}x = \cot x + c$ hence we can say the solution is false so the input will be 0.

Some important formulas used:
i.$\int {\cot x = \ln \sin x}$
ii.$y.\log (x) = \log {(x)^y}$[Logarithm power rule]
iii.$\int {{\text{cose}}{{\text{c}}^2}x = - \cot x}$
iv.${\text{cosec }}x = \dfrac{1}{{\sin x}}$

Note : Integrating factor is a function which is used to a given differential equation, it is commonly used in ordinary differential equations.
The standard form of the linear differential equation is $\dfrac{{dy}}{{dx}} + P(x)y = Q(x)$
Integrating factor of differential equation is $IF = {e^{\int {Pdx} }}$