Question

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

Answer 1

Answer:

Let the solution function be F . To show that the equation is exact you need to show that the derivative of Fx (M) with respect to y is same as derivative of Fy (N) with respect to x.

Since both derivatives are same, the equation is exact and you can find the solution by integrating either Fx or Fy .

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

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

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

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

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

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