# Solve The Given Differential Equation.

The solution of the differential equation $\frac{dy}{dx}-xtan(y-x)=1$ will be?

For solving such problems first we should see if the equation is in variable seperable form or not. Obviously here it is not. So I tried to see if it can be made to variable seperable by substitution, but substituting $y-x=z$ would not give the answer as there is one $x$ remaining outside that $tan(y-x)$. Also it is not a homogenous nor is getting converted into homogenous form so that I could substitute $y=vx$ or $x=vy$. So which method should I use here? I am getting wrong answer everytime, please help in dealing with this. Atleast provide me a hint.

• Hint: Try letting $v = y-x$. – Moo Mar 27 '17 at 4:25
• I have tried that, but the variable x in multiplication with the tan function gives me the problem. I have mentioned it in my question. – Avi Mar 27 '17 at 4:27
• You get a Separable equation as $\displaystyle \int \cot v ~.dv = \int x ~dx$. What is the problem with that? – Moo Mar 27 '17 at 4:29
• OH!! I am really, really sorry for such a silly question. I was not solving it but was just predicting the answer by looking at it. I got it when i took pen in my hand. – Avi Mar 27 '17 at 4:33

Substituting $y-x=z$ will give the solution.

let $y-x=z$ then $\dfrac{dy}{dx}=1+\dfrac{dz}{dx}$

Now, \begin{align*} \dfrac{dy}{dx}-x\cdot\tan(y-x)&=1\\ \Rightarrow1+\dfrac{dz}{dx}-x\cdot\tan z&=1\\ \Rightarrow\dfrac{dz}{dx}&=x\cdot\tan z\\ \displaystyle\int\cot z\ dz&=\displaystyle\int x\ dx+k\\ \Rightarrow\ln{\left|\sin{z}\right|}&=\dfrac{1}{2}x^2+k\\ \Rightarrow\ln{\left|\sin{z}\right|}^2&=x^2+2k\\ \Rightarrow\sin^2{(y-x)}&=e^{x^2+2k}\\ \Rightarrow\sin^2{(y-x)}&=c\cdot e^{x^2} \end{align*}

Your decision to substitute $z=y-x$ is a good one.

Differentiating with respect to $x$ gives: $$\frac{dz}{dx}=1\cdot \frac{dy}{dx}-1$$ Therefore, rearranging gives: $$\frac{dy}{dx}=\frac{dz}{dx}+1$$ Substituting gives a separable ODE: $$\frac{dz}{dx}+1-x\tan{z}=1$$ $$\frac{dz}{dx}=x\tan{z}$$ Can you continue?