Question

(2xy+y-tany)dx+(x^2-xtan^2y+sec^2y)dy=0

Answer 1

Answer:

THE QUESTION YOU ASKED MAKEN ME

Answer 2

$\sf{\left(2xy+y-\tan \left(y\right)\right)dx+\left(x^2-x\tan ^2\left(y\right)+\sec ^2\left(y\right)\right)dy=0}$

$\rule{310}{1}$

$\textsf{An ODE \sf{M(x, y)+N(x, y) y^{\prime}=0} is in exact form if the following holds :}$

$\textsf{There exists a function \sf{\Psi(x, y)} such that \sf{\Psi_{x}(x, y)=M(x, y), \quad \Psi_{y}(x, y)=N(x, y)} }$

$\textsf{ \sf{\Psi(x, y)} has continuous partial derivatives: \sf{\frac{\partial M(x, y)}{\partial y}=\frac{\partial^{2} \Psi(x, y)}{\partial y \partial x}=\frac{\partial^{2} \Psi(x, y)}{\partial x \partial y}=\frac{\partial N(x, y)}{\partial x}} }$

$\rule{310}{1}$

Let y be the dependent variable, Divide by dx :

$\to\sf{\displaystyle 2xy+y-\tan \left(y\right)+\left(x^2-x\tan ^2\left(y\right)+\sec ^2\left(y\right)\right)\frac{dy}{dx}=0}$

Substitute dy/dx with y'

$\to\sf{2xy+y-\tan \left(y\right)+\left(x^2-x\tan ^2\left(y\right)+\sec ^2\left(y\right)\right)y'\:=0}$

• The Equation is in Exact Form

$\sf{\mathsf{If\:the\:conditions\:are\:met,\:then\:}\Psi _x+\Psi _y\cdot \:y'=\frac{d\Psi \left(x,\:y\right)}{dx}=0}$

• $\sf{\mathsf{The\:general\:solution\:is\:}\Psi \left(x,\:y\right)=C}$

$\bigstar\:\:\sf{\Psi \left(x,\:y\right)=c_2}$

$\to\sf{\tan \left(y\right)+x^2y-x\left(-y+\tan \left(y\right)\right)+c_1=c_2}$

$\to\sf{\tan \left(y\right)+x^2y-x\left(-y+\tan \left(y\right)\right)=c_1}$