Question

Solve the differential equation $\dfrac{dy}{dx}=\tan \left( x+y \right)$

Answer 1

- Hint:This differential equation can be reduced to a variable separable form by putting a particular term equal to a variable. Some formulas and identities used in this question are: -
${{\sec }^{2}}t=1+{{\tan }^{2}}t;\ \int{\dfrac{dx}{x}=\ln x+c}$
\[\dfrac{d}{dx}\left( \tan t \right)={{\sec }^{2}}tdt;\ \int{\dfrac{dx}{1+{{x}^{2}}}={{\tan }^{-1}}x+c}\]

Complete step-by-step solution -

The differential equation given in the question is not of the form $f\left( x \right)dx=g\left( y \right)dy$. So, we cannot just cross multiply terms. To make it of the form $f\left( x \right)dx=g\left( y \right)dy$i.e. of the variable separable from, we have to make necessary i.e. of the variable separable from, we have to make necessary substitutions. Thus, now we will start converting into a variable separable form; in the question, it is given that,
$\dfrac{dy}{dx}=\tan \left( x+y \right)$ …………………………….(i)
Now, we will substitute $\left( x+y \right)$ to t i.e.,
$x+y=t$ ……………………………………………(ii)
We will differentiate both sides of the equation with respect to x i.e.,
$\dfrac{dx}{dx}+\dfrac{dy}{dx}=\dfrac{dt}{dx}$
$\Rightarrow 1+\dfrac{dy}{dx}=\dfrac{dt}{dx}$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{dt}{dx}-1$ …………………………………………(iii)
Now we will put the value of $\left( x+y \right)$and $\dfrac{dy}{dx}$ from equation (ii)and equation (iii) to equation (i). After doing this we will get,
$\dfrac{dt}{dx}-1=\tan t$
$\dfrac{dt}{dx}=\tan t+1$ …………………………………………(iv)
Now equation (iv) is variable separable form so we can simply cross multiply the equation. After doing this, we will get,
$\dfrac{dt}{1+\tan t}=dx$ ………………………………………….(v)
Now, we will integrate the equation (v);
$\int{\dfrac{dt}{1+\tan t}}=\int{dx}+c$ ………………………………….(vi)
Now, we will put $\tan t=p$. We will differentiate both sides. After doing this, we will get: -
${{\sec }^{2}}tdt=dp$
$\Rightarrow dt=\dfrac{dp}{{{\sec }^{2}}t}$
$\Rightarrow dt=\dfrac{dp}{1+{{\tan }^{2}}t}$ [from the identity in the hint]
$\Rightarrow dt=\dfrac{dp}{1+{{p}^{2}}}$
Putting the values in (vi), we get
$\int{\dfrac{dt}{\left( 1+{{p}^{2}} \right)\left( 1+p \right)}}=\int{dx}+c$
Now, we will separately integrate the term on LHS:
$\int{\dfrac{dt}{\left( 1+{{p}^{2}} \right)\left( 1+p \right)}}$
By using partial fraction, we get
 $\int{\dfrac{dp}{\left( 1+{{p}^{2}} \right)\left( 1+p \right)}}=\int{\dfrac{Adp}{1+p}+\int{\dfrac{\left( Bp+c \right)dp}{1+{{p}^{2}}}}}$ ……………………………(vii)
Now we have to find the values of A, B and C let us see, how this can be done: -
$\int{\dfrac{dp}{\left( 1+{{p}^{2}} \right)\left( 1+p \right)}}=\int{\dfrac{\left( A+A{{p}^{2}}+Bp+C+B{{p}^{2}}+Cp \right)dp}{\left( 1+p \right)\left( 1+{{p}^{2}} \right)}}$
So, we get; $A+C=1$
$A+B=0$
$B+C=0$
From the equations, we get: - $A=\dfrac{1}{2},B=\dfrac{-1}{2},C=\dfrac{1}{2}$
We will put these values in equation (vii). After doing this, we will get: -
$\int{\dfrac{dp}{\left( 1+{{p}^{2}} \right)\left( 1+p \right)}}=\dfrac{1}{2}\int{\dfrac{dp}{\left( 1+p \right)}+\dfrac{1}{2}\int{\dfrac{\left( -p+1 \right)dp}{\left( 1+{{p}^{2}} \right)}}}$
$=\dfrac{1}{2}\ln \left( 1+p \right)+\dfrac{1}{2}{{\tan }^{-1}}p-\dfrac{1}{4}\ln \left( 1+{{p}^{2}} \right)$
We will now put this value in the above differential equation: -
$\dfrac{1}{2}\ln \left( 1+p \right)+\dfrac{1}{2}{{\tan }^{-1}}p-\dfrac{1}{4}\ln \left( 1+{{p}^{2}} \right)=x+c$
We will now put the value back to $p=\tan t$, after doing this, we will get: -
$\dfrac{1}{2}\ln \left( 1+\tan t \right)+\dfrac{1}{2}t-\dfrac{1}{4}\ln \left( 1+{{\tan }^{2}}p \right)=x+c$
Now, we know that $t=x+y$. After substituting this, we get;
$\dfrac{1}{2}\ln \left( 1+\tan \left( x+y \right) \right)+\dfrac{1}{2}\left( x+y \right)-\dfrac{1}{4}\ln \left( 1+{{\tan }^{2}}\left( x+y \right) \right)-x=c$
$\Rightarrow \ln \left( 1+\tan \left( x+y \right) \right)+\left( y-x \right)-\dfrac{1}{2}\ln \left( 1+{{\tan }^{2}}\left( x+y \right) \right)=c$
This is the required solution of our differential equation.

Note: In the question, we are not given the values of x and y which satisfy the function. So, we cannot eliminate c from the required solution. Thus, this is a general solution where c can have any value.