Courses
Courses for Kids
Free study material
Offline Centres
More Last updated date: 28th Nov 2023
Total views: 279k
Views today: 7.79k

# Find the general solution of the following differential equation:$\sin x\dfrac{dy}{dx}+3y=\cos x$ Verified
279k+ views
Hint: To solve the given differential equation, we are going to divide $\sin x$ on both the sides of the equation. And then the equation will be written in the following differential form: $\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)$. Now, this equation will be solved by using the integrating factor concept in which first of all, we are going to find the integrating factor which is calculated as follows: $\lambda ={{e}^{\int{P\left( x \right)dx}}}$. Then we are going to use this integrating factor in this manner: $\lambda y=\int{\left( \lambda Q\left( x \right) \right)dx}$.

The differential equation given in the above problem is as follows:
$\sin x\dfrac{dy}{dx}+3y=\cos x$
Dividing $\sin x$ on both the sides of the above equation we get,
\begin{align} & \dfrac{\sin x}{\sin x}\dfrac{dy}{dx}+\dfrac{3y}{\sin x}=\dfrac{\cos x}{\sin x} \\ & \Rightarrow \dfrac{dy}{dx}+\dfrac{3y}{\sin x}=\dfrac{\cos x}{\sin x}......(1) \\ \end{align}
We know from the trigonometric conversions that:
\begin{align} & \dfrac{1}{\sin x}=\csc x; \\ & \dfrac{\cos x}{\sin x}=\cot x \\ \end{align}
Using the above relations in eq. (1) we get,
$\Rightarrow \dfrac{dy}{dx}+3y\csc x=\cot x$
The standard differential equation is as follows:
$\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)$
Now, we know the integrating factor when solving the differential equation is as follows:
$\lambda ={{e}^{\int{P\left( x \right)dx}}}$
Comparing the differential equation which we have rearranged to the above equation we get,
\begin{align} & P\left( x \right)=3\csc x; \\ & Q\left( x \right)=\cot x \\ \end{align}
Now, we are going to find the integrating factor by using the above value of P(x) in the integrating factor formula and we get,
$\lambda ={{e}^{\int{3\csc xdx}}}$ ………. (2)
We know the integration of $\csc x$ with respect to x is equal to:
$\int{\csc xdx}=\ln \left( \csc x-\cot x \right)+C$
Using the above integral in eq. (2) we get,
\begin{align} & \lambda ={{e}^{3\left( \ln \left( \csc x-\cot x \right) \right)}} \\ & \Rightarrow \lambda ={{e}^{\ln {{\left( \csc x-\cot x \right)}^{3}}}} \\ & \Rightarrow \lambda ={{\left( \csc x-\cot x \right)}^{3}} \\ \end{align}
Now, if we have an integrating factor then in the following way we are going to find the solution of the differential equation:
$\int{\left( d\left( \lambda y \right) \right)}=\int{\left( \lambda Q\left( x \right) \right)dx}$
Substituting the value of $\lambda \And Q\left( x \right)$ from the above in the above equation we get,
\begin{align} & \lambda y=\int{\left( \lambda Q\left( x \right) \right)dx} \\ & \Rightarrow {{\left( \csc x-\cot x \right)}^{3}}y=\int{\left( {{\left( \csc x-\cot x \right)}^{3}}\cot x \right)}dx \\ \end{align}
\begin{align} & \Rightarrow {{\left( \csc x-\cot x \right)}^{3}}y=\int{\left( {{\left( \dfrac{1}{\sin x}-\dfrac{\cos x}{\sin x} \right)}^{3}}\dfrac{\cos x}{\sin x} \right)}dx \\ & \Rightarrow {{\left( \csc x-\cot x \right)}^{3}}y=\int{\left( {{\left( \dfrac{1-\cos x}{\sin x} \right)}^{3}}\dfrac{\cos x}{\sin x} \right)}dx \\ \end{align}
We know that:
\begin{align} & 1-\cos x=2{{\sin }^{2}}\dfrac{x}{2}; \\ & \sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2} \\ \end{align}
Using the above identities in the above equation we get,
\begin{align} & \Rightarrow {{\left( \csc x-\cot x \right)}^{3}}y=\int{\left( {{\left( \dfrac{2{{\sin }^{2}}\dfrac{x}{2}}{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}} \right)}^{3}}\cot x \right)}dx \\ & \Rightarrow {{\left( \csc x-\cot x \right)}^{3}}y=\int{\left( {{\left( \dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}} \right)}^{3}}\dfrac{1}{\tan x} \right)}dx \\ \end{align}
We know that:
$\tan x=\dfrac{2\tan \dfrac{x}{2}}{1+{{\tan }^{2}}\dfrac{x}{2}}$
Using the above relation in the above equation we get,
\begin{align} & \Rightarrow {{\left( \csc x-\cot x \right)}^{3}}y=\int{\left( {{\left( \tan \dfrac{x}{2} \right)}^{3}}\left( \dfrac{1+{{\tan }^{2}}\dfrac{x}{2}}{2\tan \dfrac{x}{2}} \right) \right)}dx \\ & \Rightarrow {{\left( \csc x-\cot x \right)}^{3}}y=\int{\left( \dfrac{1}{2}{{\left( \tan \dfrac{x}{2} \right)}^{2}}\left( \dfrac{1+{{\tan }^{2}}\dfrac{x}{2}}{1} \right) \right)}dx \\ \end{align}
$\Rightarrow {{\left( \csc x-\cot x \right)}^{3}}y=\dfrac{1}{2}\int{\left( {{\tan }^{2}}\dfrac{x}{2}+{{\tan }^{4}}\dfrac{x}{2} \right)}dx$
We can write the above integration as:
$\Rightarrow {{\left( \csc x-\cot x \right)}^{3}}y=\dfrac{1}{2}\left( \int{\left( {{\tan }^{2}}\dfrac{x}{2} \right)dx+\int{{{\tan }^{4}}\dfrac{x}{2}dx}} \right)$
Let us take $\dfrac{x}{2}=t$ and taking derivative on both the sides of this equation we get,
\begin{align} & \dfrac{1}{2}dx=dt \\ & \Rightarrow dx=2dt \\ \end{align}
Using the above relation in the above equation we get,
$\Rightarrow {{\left( \csc x-\cot x \right)}^{3}}y=\dfrac{1}{2}\left( \int{\left( {{\tan }^{2}}dt \right)2dt+\int{{{\tan }^{4}}t2dt}} \right)$
Solving the R.H.S of the above equation we get,
$\Rightarrow {{\left( \csc x-\cot x \right)}^{3}}y=\left( \int{\left( {{\tan }^{2}}t \right)dt+\int{{{\tan }^{4}}tdt}} \right)$
We know that the integration of ${{\tan }^{2}}t$ and ${{\tan }^{4}}t$ with respect to “t” as follows:
\begin{align} & \int{{{\tan }^{2}}tdt}=\tan t-t+{{C}_{1}}; \\ & \int{{{\tan }^{4}}tdt}=\dfrac{1}{3}{{\tan }^{3}}t-\tan t+t+{{C}_{2}} \\ \end{align}
Using above integrations in the above equation we get,
$\Rightarrow {{\left( \csc x-\cot x \right)}^{3}}y=\left( \tan t-t+{{C}_{1}} \right)+\dfrac{1}{3}{{\tan }^{3}}t-\tan t+t+{{C}_{2}}$
Combining the constants ${{C}_{1}}\And {{C}_{2}}$ and writing them as C in the above equation. Also, substituting $t=\dfrac{x}{2}$ in the above equation we get,
\begin{align} & \Rightarrow {{\left( \csc x-\cot x \right)}^{3}}y=\tan \dfrac{x}{2}-\dfrac{x}{2}+\dfrac{1}{3}{{\tan }^{3}}\dfrac{x}{2}-\tan \dfrac{x}{2}+\dfrac{x}{2}+C \\ & \Rightarrow {{\left( \csc x-\cot x \right)}^{3}}y=\dfrac{1}{3}{{\tan }^{3}}\dfrac{x}{2}+C \\ \end{align}
Hence, we have solved the given differential equation.

Note: To solve the above differential equation, you must know how to use the integrating factor and then use that integrating factor in finding the final solutions of the differential equation. Also, you must know how to do the integration and along with that some trigonometric conversions. Failure of any concept will hinder your progress in this problem.