Earlier we learned about a new class of differential equations, called [b]exact differential equations[/b]. To review, a differential equation is exact if it can [i]both[/i]: be put in standard (exact) form [math]M\left(x,y\right)+N\left(x,y\right)\cdot\frac{dy}{dx}=0[/math], [i]and[/i] the functions [i]M[/i] and [i]N[/i] satisfy the partial differential equation [math]\frac{\partial}{\partial y}M\left(x,y\right)=\frac{\partial}{\partial x}N\left(x,y\right)[/math]. [br][br]After an equation is identified as being exact, the following algebraic process can be used to find the general and specific solutions. [br][br][list=1][*]Calculate [math]\Psi_M\left(x,y\right)=\int M\left(x,y\right)dx[/math] but rather than a constant of integration, add a function of integration [math]g\left(y\right)[/math][/*][*]Calculate [math]\Psi_N\left(x,y\right)=\int N\left(x,y\right)dy[/math] but rather than a constant of integration, add a function of integration [math]h\left(x\right)[/math][/*][*]Determine [math]g(y)[/math] and [math]h(x)[/math] by  reconciling [math]\Psi_M\left(x,y\right)[/math] and [math]\Psi_N\left(x,y\right)[/math] which are the same function (but viewed from two different perspectives). Call the result [math]\Psi\left(x,y\right)[/math][/*][*]Replace the symbol [math]\Psi\left(x,y\right)[/math] with a single constant of integration [i]c[/i]. [/*][*]Solve the resulting equation for [i]y [/i]to obtain the general solution.[/*][*]If an initial condition is present, then use it to solve for [i]c [/i]and obtain a specific solution.[br][/*][/list]

Try finding the specific solution of the following exact differential equation with initial condition.[br][br][math]\left(2x-y\right)+\left(2y-x\right)\cdot\frac{dy}{dx}=0;y\left(1\right)=3[/math]

[br].[br].[br].[br].[br].[br].[br]did you really give it a try? :) [br].[br].[br].[br].[br].[br].[br][br]Ok, here's the solution. [br][br]First we should check the partial differential equation condition [br][math]\frac{\partial}{\partial y}M\left(x,y\right)=\frac{\partial}{\partial x}N\left(x,y\right)[/math]. [br][br]To that end:[br][br][math]\frac{\partial}{\partial y}M\left(x,y\right)=\frac{\partial}{\partial y}\left(2x-y\right)=-1[/math][br][math]\frac{\partial}{\partial x}N\left(x,y\right)=\frac{\partial}{\partial x}\left(2y-x\right)=-1[/math][br][br]Since the partial derivatives are equal, we are now sure that the differential equation is exact, and can use the algebraic process to obtain [math]\Psi[/math].[br][br][list=1][*] [math]\Psi_M\left(x,y\right)=\int M\left(x,y\right)dx=\int2x-ydx=x^2-yx+g\left(y\right)[/math] [br][/*][*] [math]\Psi_N\left(x,y\right)=\int N\left(x,y\right)dy=\int2y-xdy=y^2-xy+h\left(x\right)[/math] [br][/*][*]By inspecting 1 and 2, [math]g\left(y\right)=y^2[/math] and [math]h\left(x\right)=x^2[/math] meaning [math]\Psi\left(x,y\right)=x^2+y^2-xy[/math][br][/*][*][math]c=x^2+y^2-xy[/math] [/*][*][math]y^2+\left(-x\right)y+\left(x^2+c\right)=0\Rightarrow y=\frac{x\pm\sqrt{x^2-4x^2+c}}{2}[/math] (do you see why? -- more work will be done in class to explain this step)[/*][*][math]y\left(1\right)=3=\frac{1\pm\sqrt{-3+c}}{2}\Rightarrow c=28[/math] (do you see why? -- more work will be done in class to explain this step)[/*][/list]Therefore the specific solution is[br][br][math]y\left(x\right)=\frac{x\pm\sqrt{x^2-4x^2+28}}{2}[/math][br][br]Which can be tidied up as[br][br][math]y\left(x\right)=\frac{x\pm\sqrt{-3x^2+28}}{2}[/math][br][br]The slope field of the equation, the specific solution, and the general solution are visualized below. Note that the slope field form of the equation, as always for exact equations is, [math]\frac{dy}{dx}=-\frac{M\left(x,y\right)}{N\left(x,y\right)}=-\frac{2x-y}{2y-x}[/math]

Try solving this exact equation.[br][br][math]\left(9x^2+y-1\right)-\left(4y-x\right)\cdot\frac{dy}{dx}=0;y\left(0\right)=1[/math][br][br]Hint: the above equation is exact (but you should still check it), but it is NOT in standard (exact) form.[br][br]After you are done solving this exact differential equation, try using GeoGebra to plot the slope field, the initial condition, and check your specific solution in the blank GeoGebra window below.

[br].[br].[br].[br].[br].[br].[br]did you really give it a try? :) [br].[br].[br].[br].[br].[br].[br][br]Ok, here's the solution. [br][br]First, we need to put it in standard (exact) form by incorporating the negative sign inside the parentheses:[br][br][math]\left(9x^2+y-1\right)+\left(-4y+x\right)\cdot\frac{dy}{dx}=0;y\left(0\right)=1[/math][br][br]First we should check the partial differential equation condition [br][math]\frac{\partial}{\partial y}M\left(x,y\right)=\frac{\partial}{\partial x}N\left(x,y\right)[/math]. [br][br]To that end:[br][br][math]\frac{\partial}{\partial y}M\left(x,y\right)=\frac{\partial}{\partial y}\left(9x^2+y-1\right)=1[/math][br][math]\frac{\partial}{\partial x}N\left(x,y\right)=\frac{\partial}{\partial x}\left(-4y+x\right)=1[/math][br][br]Since the partial derivatives are equal, we are now sure that the differential equation is exact, and can use the algebraic process to obtain [math]\Psi[/math].[br][br][list=1][*] [math]\Psi_M\left(x,y\right)=\int M\left(x,y\right)dx=\int9x^2+y-1dx=3x^3+yx-x+g\left(y\right)[/math] [br][/*][*] [math]\Psi_N\left(x,y\right)=\int N\left(x,y\right)dy=\int-4y+xdy=-2y^2+xy+h\left(x\right)[/math] [br][/*][*]By inspecting 1 and 2, [math]g\left(y\right)=-2y^2[/math] and [math]h\left(x\right)=3x^3-x[/math] meaning [math]\Psi\left(x,y\right)=3x^3+yx-x-2y^2[/math][br][/*][*][math]c=3x^3+yx-x-2y^2[/math] [/*][*]This is quadratic in y, so put it in standard (quadratic form) and then apply the quadratic formula:[math]\left(2\right)y^2+\left(-x\right)y+\left(-3x^3+x+c\right)=0\Rightarrow y=\frac{x\pm\sqrt{x^2-4\left(2\right)\left(-3x^3+x+c\right)}}{4}[/math]. Distributing the -8 over A*C leads to the simplification [math]y=\frac{x\pm\sqrt{x^2+24x^3-8x+c}}{4}[/math][/*][*][math]y\left(0\right)=1=\frac{0\pm\sqrt{0+0-0+c}}{4}\Rightarrow c=16[/math] [/*][/list]Therefore the specific solution is[br][br][math]y\left(x\right)=\frac{x\pm\sqrt{x^2+24x^3-8x+16}}{4}[/math][br][br]The slope field of the equation, the specific solution, and the general solution are visualized below. Note that the slope field form of the equation, as always for exact equations is, [math]\frac{dy}{dx}=-\frac{M\left(x,y\right)}{N\left(x,y\right)}=-\frac{9x^2+y-1}{-4y+x}[/math]