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 \[IF = {e^{\int {Pdx} }}\]