Question

The solution of the differential equation $\dfrac{y+\sin x{{\cos }^{2}}\left( xy \right)}{{{\cos }^{2}}\left( xy \right)}dx+\left( \dfrac{x}{{{\cos }^{2}}\left( xy \right)}+\sin y \right)dy=0$ is
$\left( a \right)\text{ }\tan xy+\cos x+\cos y=c$ $\left( b \right)\text{ }\tan xy-\cos x-\cos y=c$
$\left( c \right)\text{ }\tan xy+\cos x-\cos y=c$ $\left( d \right)$ None of these

Answer 1

To get the solution of the given differential equation, we will multiply by ${{\cos }^{2}}\left( xy \right)$ both sides of the given differential equation. Then, we will get the equation in simple form in the form of $x$ and $y$ . After that, we will apply an integration method so that we will have the final solution of the given differential equation.

Complete step-by-step solution:
Since, the given differential equation is as:
$\Rightarrow \dfrac{y+\sin x{{\cos }^{2}}\left( xy \right)}{{{\cos }^{2}}\left( xy \right)}dx+\left( \dfrac{x}{{{\cos }^{2}}\left( xy \right)}+\sin y \right)dy=0$
Now, we will multiply by ${{\cos }^{2}}\left( xy \right)$in the above differential equation so that the above differential equation will be easy to proceed as:
$\Rightarrow \left[ \dfrac{y+\sin x{{\cos }^{2}}\left( xy \right)}{{{\cos }^{2}}\left( xy \right)}dx+\left( \dfrac{x}{{{\cos }^{2}}\left( xy \right)}+\sin y \right)dy \right]\times {{\cos }^{2}}\left( xy \right)=0\times {{\cos }^{2}}\left( xy \right)$
Here, we will open the bracket so that we can multiply by ${{\cos }^{2}}\left( xy \right)$ in every term of above differential equation and multiplication of any number with zero is always equal to zero. So, we will have the above equation as:
$\Rightarrow \left[ \left( \dfrac{y+\sin x{{\cos }^{2}}\left( xy \right)}{{{\cos }^{2}}\left( xy \right)}\times {{\cos }^{2}}\left( xy \right) \right)dx+\left( \dfrac{x}{{{\cos }^{2}}\left( xy \right)}\times {{\cos }^{2}}\left( xy \right)+\sin y\times {{\cos }^{2}}\left( xy \right) \right)dy \right]=0\times {{\cos }^{2}}\left( xy \right)$Since, we have applied the multiplication in every term in the above differential equation by ${{\cos }^{2}}\left( xy \right)$ , we have the above differential equation as:
\Rightarrow \left[ \left\\{ y+\sin x{{\cos }^{2}}\left( xy \right) \right\\}dx+\left\\{ x+\sin y{{\cos }^{2}}\left( xy \right) \right\\}dy \right]=0
Now, we will open the bracket completely. So, we will have the above differential equation as:
$\Rightarrow ydx+\sin x{{\cos }^{2}}\left( xy \right)dx+xdy+\sin y{{\cos }^{2}}\left( xy \right)dy=0$
Here, we will rearrange the above differential equation as:
$\Rightarrow xdy+ydx+\sin x{{\cos }^{2}}\left( xy \right)dx+\sin y{{\cos }^{2}}\left( xy \right)dy=0$ … $\left( i \right)$
Since, we know that the differentiation of multiplication term is as:
$\Rightarrow d\left( u.v \right)=u.dv+v.du$
Similarly we can write for $x$ and $y$ as:
$\Rightarrow d\left( x.y \right)=x.dy+y.dx$ … $\left( ii \right)$
Now, we can replace $x.dy+y.dx$ by $d\left( x.y \right)$ by using equation $\left( ii \right)$ in equation $\left( i \right)$ as:
$\Rightarrow d\left( xy \right)+\sin x{{\cos }^{2}}\left( xy \right)dx+\sin y{{\cos }^{2}}\left( xy \right)dy=0$
Now, we have common element two terms, we can write the above equation as:
$\Rightarrow d\left( xy \right)+\left( \sin xdx+\sin ydy \right){{\cos }^{2}}\left( xy \right)=0$
Here, we will divide the above differential equation by ${{\cos }^{2}}\left( xy \right)$ as:
$\Rightarrow \dfrac{d\left( xy \right)}{{{\cos }^{2}}\left( xy \right)}+\dfrac{\left( \sin xdx+\sin ydy \right){{\cos }^{2}}\left( xy \right)}{{{\cos }^{2}}\left( xy \right)}=0$
Since, ${{\cos }^{2}}\left( xy \right)$ will cancel out ${{\cos }^{2}}\left( xy \right)$ in second term. So, the above equation can be written as:
$\Rightarrow \dfrac{1}{{{\cos }^{2}}\left( xy \right)}d\left( xy \right)+\sin xdx+\sin ydy=0$
Although, we know that $\dfrac{1}{{{\cos }^{2}}\left( xy \right)}$ is equal to ${{\sec }^{2}}\left( xy \right)$. Here, we will use it in the above differential equation. So, the above differential equation will be as:
$\Rightarrow {{\sec }^{2}}\left( xy \right)d\left( xy \right)+\sin xdx+\sin ydy=0$
Now, we will apply integration sign in all the terms of above equation, so the above equation would be as:
$\Rightarrow \int{{{\sec }^{2}}\left( xy \right)}d\left( xy \right)+\int{\sin x}dx+\int{\sin y}dy=0$
Here, we will integrate the every term of above differential equation and will write integration value of the above differential equation as:
$\Rightarrow \tan \left( xy \right)-\cos x-\cos y+c=0$
Since, $c$ is an unknown constant. So, we can change its place and can write the above equation as:
$\Rightarrow \tan \left( xy \right)-\cos x-\cos y=c$
Hence, option $\left( b \right)$ shows the write answer of the solution of the given differential equation.

Note: Since, we got the solution from the given differential equation. Now, we will check from the solution whether we are able to find the given differential equation or not in the following way as:
Here, the obtained answer that is:
$\Rightarrow \tan \left( xy \right)-\cos x-\cos y=c$
Now, we will differentiate it as:
$\Rightarrow d\left( \tan \left( xy \right)-\cos x-\cos y \right)=dc$
Here, we will open the bracket and will apply differentiation sign separately as:
$\Rightarrow d\tan \left( xy \right)-d\cos x-d\cos y=dc$
As we know that the differentiation of a constant is always zero and in the differentiation of double variable, second variable will be differentiate again as:
$\Rightarrow {{\sec }^{2}}\left( xy \right)d\left( xy \right)-\left( -\sin xdx \right)-\left( -\sin ydy \right)=0$
Here, we will solve the differentiation of second variable in first term and will open the bracket for all other terms as:
$\Rightarrow {{\sec }^{2}}\left( xy \right)\left( xdy+ydx \right)+\sin xdx+\sin ydy=0$
Since, we know that ${{\sec }^{2}}\left( xy \right)$ can be written as $\dfrac{1}{{{\cos }^{2}}\left( xy \right)}$ . So, the above equation will be as:
$\Rightarrow \dfrac{1}{{{\cos }^{2}}\left( xy \right)}\left( xdy+ydx \right)+\sin xdx+\sin ydy=0$
Now, we will apply multiplication method in the above equation as:
$\Rightarrow \dfrac{x}{{{\cos }^{2}}\left( xy \right)}dy+\dfrac{y}{{{\cos }^{2}}\left( xy \right)}dx+\sin xdx+\sin ydy=0$
Here, we will combine equal like terms and will make its common factor as:
$\Rightarrow \dfrac{y}{{{\cos }^{2}}\left( xy \right)}dx+\sin xdx+\sin ydy+\dfrac{x}{{{\cos }^{2}}\left( xy \right)}dy=0$
$\Rightarrow \dfrac{y}{{{\cos }^{2}}\left( xy \right)}dx+\sin xdx+\left( \sin y+\dfrac{x}{{{\cos }^{2}}\left( xy \right)} \right)dy=0$
We can write the above equation as:
$\Rightarrow \dfrac{y+\sin x{{\cos }^{2}}\left( xy \right)}{{{\cos }^{2}}\left( xy \right)}dx+\left( \sin y+\dfrac{x}{{{\cos }^{2}}\left( xy \right)} \right)dy=0$
Since, we got the question from the obtained answer. Hence, the solution is correct.