Differential Equation Notes PDF

# TABLE OF CONTENTS ## I. INTRODUCTION TO DIFFERENTIAL EQUATIONS (D.E.) * **A. TYPES OF DIFFERENTIAL EQUATIONS** * **B. ORDER OF DIFFERENTIAL EQUATIONS** * **C. DEGREE OF DIFFERENTIAL EQUATIONS** * **D. BASIC SOLUTIONS TO D.E.** ## II. SOLUTIONS TO DIFFERENTIAL EQUATIONS * **A. ELIMINATION OF AR...

# TABLE OF CONTENTS ## I. INTRODUCTION TO DIFFERENTIAL EQUATIONS (D.E.) * **A. TYPES OF DIFFERENTIAL EQUATIONS** * **B. ORDER OF DIFFERENTIAL EQUATIONS** * **C. DEGREE OF DIFFERENTIAL EQUATIONS** * **D. BASIC SOLUTIONS TO D.E.** ## II. SOLUTIONS TO DIFFERENTIAL EQUATIONS * **A. ELIMINATION OF ARBITRARY CONSTANT** * **B. ELIMINATION OF ARBITRARY CONSTANT USING ELIMINATION METHOD** * **C. FAMILIES OF CURVE** * **D. VARIABLE SEPARABLE D.E.** * **E. HOMOGENEOUS D.E.** * **F. EXACT DE.** * **G. NON-EXACT D.E.** * **H. FIRST ORDER LINEAR D.E.** * **I. BERNOULLIS EQUATION** ## III. APPLICATION OF FIRST ORDER D.E. * **A. LAWS OF GROWTH & DECAY** * **B. NEWTON'S LAW OF COOLING** ## IV. APPLICATION OF FIRST ORDER D.E. IN ELECTRICAL CIRCUITS * **A. RL CIRCUITS** * **B. RC CIRCUITS** ## V. HOMOGENEOUS D.E. W/CONSTANT COEFFICIENTS * **A. AUXILIARY EQUATION W/DISTINCT ROOTS** * **B. AUXILIARY EQUATION W/REPEATED ROOTS** ## VI. LAPLACE TRANSFORM INTRODUCTION IN ADVANCE MATHEMATICS * **A. LAPLACE TRANSFORM INTRODUCTION** * **B. LAPLACE TRANSFORM OF SINEX COSINE FUNCTION** * **C. LAPLACE TRANSFORM OF EXPONENTIAL FUNCTION** # **DIFFERENTIAL EQUATIONS** * Is a mathematical equation that relates some function with its derivatives. ## **ORDINARY D.E.** * Is a D.E. containing one or more functions of one independent variable, and has derivatives of those functions. ## **PARTIAL D.E.** * Is a **D.E.** that involves two or more independent variables, an unknown function (dependent on those variables) and partial derivatives of unknown function. ## **ORDER OF D.E.** * It is the highest/largest derivative present in the D.E. ## **DEGREE OF D.E.** * It is the exponent at which the order is raised. ## **EXAMPLES:** 1. $ay'' + by' + cy = g(t)$ * ODE, 2nd order, 1st degree 2. $\frac{dy}{dx} = (-y) + y^2e^y$ * ODE, 2nd order, 1st degree. 3. $y^2 + 10y''' - 4y' + 2y = cos(t)$ * ODE, 4th order, 1st degree 4. $\frac{d^2x}{dt^2} + k^2x = 0$ * ODE, 2nd order, 1st degree 5. $(\sqrt{\frac{dy}{dx}} + 3y)\frac{d^2y}{dx^2} = (\frac{d^2y}{dx^2} + 3y)^2$ * ODE, 2nd order, 2nd degree 6. $(1 + (\frac{dy}{dx})^2)^2 = (1+ \frac{d^2y}{dx^2} - y (\frac{d^2y}{dx^2})^2)^2$ * ODE, 3rd order, 4th degree 7. $\frac{d^2y}{dx^2} + k^2y = c + g(t)$ * PDE, 2nd order, 1st degree ## **SOLUTIONS TO D.E.** 1. **Verify** that $y(x) = \frac{x}{k}$ is a solution to the D.E. $4x^2y'' + 2xy' + 3y=0$. * **Solution:** * $y(x)=x^2$, * $4xy'' + 12xy' + 3y= 0$ * $y'= \frac{1}{2}x^\frac{-3}{2}$ * $y''= \frac{-5}{4}x^\frac{-7}{2}$ * Substitute to the Equation. * $4x^2(\frac{-5}{4}x^\frac{-7}{2}) + 12x(\frac{1}{2}x^\frac{-3}{2}) +3(x^2) = 0$ * $-\frac{5}{2}x^\frac{-3}{2}+ 6x x^\frac{-3}{2} + 3x^2 = 0$ * $0 = 0$ 2. **Verify** that $e^{2x} = y$ is a solution to the equation $y'''- 3y' + 2y = 0$. * **Solution:** * $y = e^{2x}$ * $y' = 2e^{2x}$ * $y'' = 4e^{2x}$ * $y'''= 8e^{2x}$ * Substitute to Equation: * $8e^{2x} - 3(2e^{2x}) + 2e^{2x} = 0$ * $0 = 0$ 3. $Sinkt$; $\frac{d^2x}{dt^2} + k^2x = 0$ * *Solution:* * Substitute to the Equation. * $x = sinkt$ * $\frac{d^2x}{dt^2} = -ksinkt$ and $\frac{dx}{dt} = kcoskt$ * $-ksinkt + k^2sinkt = 0$ * $0 = 0$ 4. $y= Cie^{-2x} + C_2e^{3x}$ * *Solution:* * $y' = -2C_1e^{-2x} + 3C_2e^{3x}$ * $y'' = 4C_1e^{-2x} + 9C_2e^{3x}$ * $y''' = -8C_1e^{-2x} + 27C_2e^{3x}$ 5. $x = C_1cos(\omega t) + C_2sin(\omega t)$ * *Solution:* * $\frac{dx}{dt} = -C_1\omega sin(\omega t) + C_2\omega cos(\omega t)$ * $\frac{d^2x}{dt^2} = -C_1\omega^2cos(\omega t) - C_2\omega^2sin(\omega t)$ * $\frac{d^3x}{dt^3} = C_1\omega^3sin(\omega t) - C_2\omega^3cos(\omega t)$ * $\frac{d^4x}{dt^4} = C_1\omega^4cos(\omega t) + C_2\omega^4sin(\omega t)$ * $\frac{d^4x}{dt^4} = \omega^4(C_1cos(\omega t) + C_2sin(\omega t))$ * $\frac{d^4x}{dt^4} = \omega^4x = 0$ # **ELIMINATION OF ARBITRARY CONSTANT** * The Elimination of Arbitrary Constant is necessary to obtain the differential equation. ## **EXAMPLES:** 1. $ax^2 + bx + cy = 0$ 2. $2ax + b = y'$ 3. $2a = y''$ 4. $y'=0$ * **The** rule is, we will differentiate the given equation based on the number of arbitrary constants. # **ELIMINATION OF ARBITRARY CONSTANT USING ELIMINATION METHOD (DETERMINANT)** 1. $y = C_1 + C_2e^{3x}$ * $y' = 0 + 3C_2e^{3x}$ * $y'' = 0 + 9C_2e^{3x}$ * $-y + C_1 + C_2e^{3x} = 0$ * $-y' + 0 + 3C_2e^{3x} = 0$ * $-y'' + 0 + 9C_2e^{3x} = 0$ * $\begin{vmatrix} -y & 1 & e^{3x} \\ -y' & 0 & 3e^{3x} \\ -y'' & 0 & 9e^{3x} \end{vmatrix} = 0$ * $0 = (0) - b(-y) + (0) - (0 + 0 - ay)$ * $[0 - -3y^2 + 9y’] / \frac{1}{2}$ * $[0 - -y + 3y’] - 1$ * $y'' - 3y' = 0$ 2. $y = C_1e^{2x} + C_2e^{3x}$ * $y' = 2 C_1e^{2x} + 3C_2e^{3x}$ * $y'' = 4C_1e^{2x} + 9C_2e^{3x}$ * $-y + C_1e^{2x} + C_2e^{3x} = 0$ * $-y' - 2C_1e^{2x} + 3C_2e^{3x} = 0$ * $-y'' + 4C_1e^{2x} + 9C_2e^{3x} = 0$ * $\begin{vmatrix} -y & 1 & 1 \\ -y' & 2 & 3 \\ -y'' & 4 & 9 \end{vmatrix} = 0$ * 0 = $y'' - 3y' = 0$ # **FAMILIES OF CURVES** * Are equations that includes arbitrary constants. This equation are mostly standard equations of conic sections of lines, circles, parabolas, ellipse k hyperbolas. ## **EXAMPLES:** 1. **Find** the D.E. satisfied by the families of line with slope and y-intercept equal. * **Solution:** * $y = mx + b$, $m=b$ * $y= mx+m$ * $y'=m$ * $y= y'x+y'$ * $y' y+(y')^2 = 0$ * $y' (x+1) - y = 0$ * $\frac{dy}{dx}(x+1)-y=0$ * $(x+1)dy-ydx = 0$ 2. **Circles** w/ origin at center * **Solution:** * $x^2+y^2=r^2$ * $x^2 + y(y') = 0$ * $x+yy' = 0$ * $dx + yy'dx = 0$ * $xdx +ydy=0$ 3. **Circles** w/ center at x-axis * **Solution:** * $(x-h)^2 + y^2 = r^2$ * $2(x-h)(1) + 2yy' = 0$ * $x-h+ yy' = 0$ * $x + yy' = h$ * $1 + (y')(y'') + (y')(y') = 0$ * $1+yy'' + (y')^2 = 0$. 4. **Straight** lines w/ slope y, x-intercept equal. * **Solution:** * $y = mx + b$ * $o = mx + b$ * $-b = mx$ * $if-x=m$ * $m=-b$ * $m^2 = b$ * $b=m²-m$ * $y = mx - m^2$ * $y'-m = 0$ * $y=y'x-(y')^2$ * $(y')^2 -y'y +y=0$ 5. **Circles** w/ center at y-axis * **Solution:** * $x^2 + (y-k)^2 = r^2$ * $2x + 2(y-k)y' = 0$ * $x+(y-k)y'= 0$ * $x + yy' - ky'=0$ * $x + yy' - ky' = 0$ * $y'(x+yy'' + y'y') - (x+yy')(y'') = 0$ * $\frac{y'}{y}\Bigg(x+yy'' + y'y' - \frac{(x+yy')(y'')}{y'}\Bigg) = 0$ * $-xy'' + (y')^3 + y' = 0$ 5. **Family** of parabolas having their vertices at origin and foci on y-axis * **Solution:** * $4ay=x^2; 4a=a$ * $ay=x^2$ * $ay' = 2x$ * $y' = \frac{2x}{y}$ * $2y = \frac{x^2}{y}$ * $2y^2=x^2$ * $x(\frac{dy}{dx}) -2y = 0$ * $xdy-2ydx = 0$ 6. **Family** of parabolas w/ vertex on y-axis and parallel to y-axis. * **Solution:** * $(x-h)^2=4ay = ay$ * $2(x-h) \frac{dy}{dx} = ay$ * $2x -2h = ay'$ * $2x = ay' + 2h$ * $\frac{2}{y'} = \frac{ay'}{2x}$ * $2x - 2h = \frac{2y}{y'}$ * $x-h = \frac{y}{y'}$ * $(x-h)^2 = (\frac{y}{y'})^2$ * $(x-h)^2 = \frac{y}{y'}(\frac{y}{y'}) + \frac{y}{y'}\Bigg(\frac{y}{y'}\Bigg)'$ * $x^2 - 2x(x-1) + (x-1)^2 = \frac{y}{y'}[2yy'' + (y')^2]$ * $x^2 + 2x - x^2 + 2x = \frac{yy''}{y'} + \frac{y}{y'} (y')^2$ * $x^2 + 2x = (\frac{y}{y'})^2 + y^2$ * $2yy' - (y')^2 = 0$. # **VARIABLE SEPARABLE DIFFERENTIAL EQUATION** * All terms containing x & y must be grouped with dx and dy respectively. If we can group them by dx and py, the D.E. is said to be variable separable. ## **TYPES OF SOLUTION:** 1. **General** solution * A solution that contains, however, the arbitrary constant does not yet defined. 2. **Particular** solution * A solution that no longer contains arbitrary constant. Hence, the arbitrary constant are defined. ## **EXAMPLES:** 1. $y=2x$ * **Solution:** * $\frac{dy}{dx}=2y$ * $\frac{1}{2y}dy = dx$ * $\frac{1}{2}\int \frac{1 }{y}dy =\int dx$ * $\frac{1}{2}[ln(y) =ln(x)+c]2$ * $ln(y) = 2ln(x)+c$ * $ln(y) = ln(x)^2 + c$ * $e^{ln(y)} = e^{ln(x)^2} + c$ * $y = xe^c$ but $e^c=c$ * $y = xe^c$ 2. $2x(y+1)dx-ydy=0$ * **Solution:** * $@x = 1, y= 0$ * $2xdx-ydy = 0$ * $\int 2xdx = \int ydy$ * $2\int xdx = \int \frac{y}{y+1}dy$ * LET: $du = dy$ * $\int du = \int \frac{y}{y+1}dy$ * $u = y + 1$ * $x^2 = \int \frac{y+1-1}{y+1} dy$ * $x^2 = \int dy - \int\frac{1}{y+1}dy$ * $x^2 = y - ln(y+1) + c$ * $@x=1, y = 0$ * $(1)^2 = 0 - ln(0+1) + c$ * $x^2 - y -ln(y+1) + 1 = c$ 3. $2xyy' = 1 + y^2$; $@x = 2, y = 3$ * **Solution:** * $2xy(\frac{dy}{dx}) = 1+y^2$ * $2xydy = (1 + y^2)dx$ * $2xydy = (1 + y^2)dx$ * $[2xydy = (1 + y^2)dx] / 2x(1 + y^2)$ * $\frac{ydy}{1 + y^2} = \frac{dx}{2x}$ * $\int \frac{ydy}{1+y^2} = \int\frac{dx}{2x} $ * LET : $du = 2ydy$ * $\int \frac{du}{u} = \frac{1}{2} \int \frac{dx}{x}$ * $\frac{1}{2}ln(u) = \frac{1}{2}ln(x)+c$ * $\frac{1}{2}ln(y^2 + 1) = \frac{1}{2}ln(x)+c$ * $[ln(y^2 + 1) = ln(x) + c ] 2$ * $ln(y^2 + 1) = ln(x)^2 + c$ * $e^{ln(y^2 + 1)} = e^{ln(x)^2 + c}$ * $y^2 + 1 = e^{ln(x)^2} e^c$ * $y^2 + 1 = Cx$ * $@ x = 2, y = 3$ * $(3)^2 +1 = 2C$ * $C = 5$ * $y^2+ 1 = 5x$ 4. $(x^3 + 2x)y dx - x(y^4 + 3)y dy = 0$ * **Solution:** * $(x^3 + 2x)ydx = x(y^4 + 3)y dy$ * $[(x^3 + 2x)y dx = x^(y^4 + 3)y dy]/yx$ * $\int \frac{x^3 + 2x}{x}dx = \int \frac{y^4+3}{y}dy$ * $\int x^2dx + 2 \int dx = \int y^3 dy + 3\int \frac{1}{y}dy$ * $\int x^2 dx + 2 \int dx = \int y^3dy + 3 \int \frac{1}{y}dy$ * $\frac{1}{3}x^3 + 2ln(x) = \frac{1}{4}y^4 + 3ln(y) + c$ * $@ x = 1, y = 1$ * $(1)^3 + 2ln(1) = \frac{1}{4}(1)^4 + 3ln(1) + c$ * $\frac{1}{3} + c = \frac{1}{4} + c$ * $\frac{1}{3} + c \rightarrow 0$ * $\frac{1}{3}x^3 + 2ln(x) = \frac{1}{4}y^4 + 3ln(y) + \frac{1}{12}$ 5. $ye^{2x}dx = (4+e^{2x})dy$ * **Solution:** * $ye^{2x}dx = (4+e^{2x})dy$ * $ye^{2x}dx = (4+e^{2x})dy$ * $\frac{e^{2x}dx}{4 + e^{2x}} = \frac{dy}{y}$ * LET: $du = 4 + e^{2x}$ * $du = 2e^{2x}dx$ * $\frac{1}{2}\int \frac{du}{u} = \int \frac{dy}{y}$ * $\frac{1}{2}ln(u) = ln(y)+c$ * $[\frac{1}{2}ln(4 + e^{2x})= ln(y) + c]$ * $ln(4 + e^{2x}) = 2ln(y) + c$ * $ln(4 + e^{2x}) = ln(y)^2 + c$ * $e^{ln(4+e^{2x})} = e^{ln(y)^2 + c}$ * $4 + e^{2x} = Cy^2$ # **HOMOGENEOUS DIFFERENTIAL EQUATIONS** * A **differential** equation is said to be a **homogeneous D.E.** if every **term** has the **same degree**. * Homogeneous can be written **as** $M(x, y)dx + N(x, y)dy = 0$ * If M is simpler, **use:** * $x = vy, dx = vdy + ydx$ * If N is simpler **use:** * $y = vx, dy = vdx + xdv$ ## **EXAMPLES:** 1. $(x^2 - xy + y^2)dx - xy^2 dy = 0$ * **Solution:** * Let $y = vx, dy = vdx + xdv$ * $(x^2 - x(vx) + (vx)^2)dx - x(vx)^2(vdx + xdv) = 0$ * $(x^2 - vx^2 + v^2x^2)dx - (x^2)(vdx + vdv) = 0$ * $(x^2 - vx^2 + v^2x^2)dx - v^2x^2dx - v^2xdx = 0$ * $v^2xdx=0$ * $x^2dx - x^2dx - v^2dx + v^2xdx = 0$ * $v^2dx = 0$ * $dv = 0$ * $\int dv = \int 0$ * $\int \frac{dv}{v} = \int 0$ * Let: $du=-dv$ * $lnx = \int \frac{dv}{v} + \int \frac{-du}{u}$ * $lnx = \int \frac{dv}{v} + \int \frac{du}{u}$ * $lnx = - In (1-(1/v)) + ln(u) + c$ * $lnx = - In(1 - \frac{y}{x}) + ln(-y{x}) + c$ * $lnx = In(\frac{x}{x-y}) - In(\frac{y}{x}) + c$ 2. $xydx + (x^2 + y^2)dy = 0$ * **Solution:** * Let $x = vy, dx = vdy + ydx$ * $vy(vdy + ydx) + (v^2y^2 + y^2)dy = 0$ * $vy^2(vdy + ydx) + v^2y^2dy + y^2dy = 0$ * $v^2y^2dy + vy^3dv + v^2y^2dy + y^2dy=0$ * $[2v^2y^2dy + vy^3dv + y^2dy = 0]/ y^2$ * $2v^2dy + vydv + dy = 0$ * $2v^2dy + dy + vydv = 0$ * $2v^2dy + dy = - vdy$ * $\int \frac{2v^2dy + dy}{2v^2 +1} = \int -vdv$ * LET: $u = 2v^2 + 1$ * $du = 4vdv$ * $\int\frac{du}{4u} = \int -vdv$ * $\frac{1}{4}\int \frac{du}{u} = \int -vdv$ * $In y + \frac{1}{4} \int \frac{du}{u} = 0$ * $In y + \frac{1}{4} ln(2v^2 +1) = 0$ * $[In y +\frac{1}{4} ln(2v^2 + 1) = 0] 4$ * $4In y + ln(2v^2 + 1) = c$ * $In y^4 + ln(2v^2 + 1) = c$ * $In y^4 + (2v^2 +1) = c$ * $e^{In y^4 + (2v^2 + 1)} = e^c$ * $y^4(2v^2 + 1) = c$ * LET: $v=\frac{x}{y}$ * $y^4(2(\frac{x}{y})^2 + 1) = c$ * $y^2(2\frac{x^2}{y^2} + 1) = c$ * $y^2(\frac{2x^2 + y^2}{y^2}) = c$ * $2x^2 + y^2 = c$ * WHEN $x=0, y = 1$ * $(1)^2(2(0)^2 + (1)^2) = c$ * $1 = c$ * $y^2(2x^2 + y^2) = 1$ # **EXACT DIFFERENTIAL EQUATION** * A D.E. is said to be exact if: * $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ * **TEST** FOR EXACTNESS (T.F.E) * To solve for the exact d.e., we need to use the **TFE** formulas: * $\frac{\partial F}{\partial x} = M(x, y) < \frac{\partial F}{\partial y} = N(x, y)$ * **WHERE** F is the solution to an exact D.E. ## **EXAMPLES:** 1. $(3x^2y-6x) dx+ (x^2 + 2y)dy = 0$ * **Solution:** * $M = 3x^2y-6x$ * $N = x^2 + 2y$ * $\frac{\partial M}{\partial y} = \frac{d}{dy}(3x^2y - 6x) = 3x^2$ * $\frac{\partial N}{\partial x}= \frac{d}{dx}(x^2 + 2y) = 2x +0 = 2x$ * $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ .. **EXACT** * **dF** = $\int (3x^2y-6x)dx$ * F = $\int 3x^2ydx - \int 6xdx$ * F = $3yxdx - 6 \int xdx$ * F= $y(\frac{x^3}{3}) - 6(\frac{x^2}{2}) + Q(y)$ * F = $x^2y - x^2 + Q(y)$ * $\frac{\partial F}{\partial y} = N(x, y) \rightarrow \frac{\partial F}{\partial y} = x^2 + 2y$ * dF = $\int(x^2 +2y)dy$ * F = $\int x^2 dy + \int 2ydy$ * F = $x^2 \int dy + 2 \int ydy$ * F = $x^2y + 2(\frac{y^2}{2}) + R(x)$ * F = $x^2y + y^2 + R(x)$ * F = $x^2y + y^2 + R(x)$ * $x^2y + y^2 + R(x) = x^2y-3x^2+ Q(y)$ * $R(x) = -3x^2$ * $Q(y) = y^2$ * $x^2y + y^2 - 3x^2 = C$ OR * $x^2y - 3x^2 + y^2 = C$ 2. $(2x^3 - xy^2 - 2y + 3)dx - (x^2y + 2x)dy = 0$ * *Solution:* * M = $(2x^3 - xy^2 - 2y + 3)$ * N = $ - (x^2y + 2x)$ * $\frac{\partial M}{\partial y} = \frac{d}{dy}(2x^3-xy^2-2y +3) = (0 + (-2xy) - 2 + 0) = -2xy - 2$ * $\frac{\partial N}{\partial x}= \frac{d}{dx}(x^2y + 2x) = (2x-2) = - 2xy - 2$ * $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ ... **EXACT D.E.** * **dF**= $\int(2x^3 - xy^2 - 2y +3)dx$ * F= $\int 2x^3dx - \int xy^2 dx - \int2ydx + 3 \int dx$ * F= $2\int x^3dx - y^2\int xdx - 2y \int dx + 3 \int dx$ * F= $2(\frac{x^4}{4})-y^2(\frac{x^2}{2}) -2y(x) + 3(x) + Q(y)$ * F= $2(\frac{x^4}{4}) -y^2(\frac{x^2}{2}) -2yx + 3x + Q(y)$ * $\frac{\partial F}{\partial y} = N(x, y) \rightarrow \frac{\partial F}{\partial y} = (x^2y + 2x)$ * *dF = $\int (-x^2y - 2x)dy$* * F = $\int -x^2ydy - \int 2xdy$ * F = $-x^2\int ydy - 2x\int dy$ * F = $-x^2(\frac{y^2}{2}) - 2xy +R(x)$ * F = $-x^2(\frac {y^2}{2}) -2xy + 3x + Q(y) = -x^2(\frac{y^2}{2}) -2xy + R(x)$ * $Q(y) = 0$; $R(x) = 3x$ * C = $x^2 - x^2y^2 - 4xy + 6x $ # **NON-EXACT D.E. WITH SPECIAL INTEGRATING FACTORS** * If $(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})/N = f(x)$. Then integrating factor: * $I.F. = e^{\int f(x)dx}$ * If $(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})/M = f(y)$. Then integrating factor is slightly different: * $I.F. = e^{\int f(y)dy}$ ## **EXAMPLES:** 1. $(y - xy)dx + xdy = 0$ * **Solution:** * $M = (y - xy)$ * $N = x$ * $\frac{\partial M}{\partial y} = \frac{d}{dy}(y-xy) = 1-x$ * $\frac{\partial N}{\partial x} = \frac{d}{dx}(x) = -1$ * $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$ ... **NOT EXACT** * $(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})/N = f(x)$ * $\frac{(1-x)-(-1)}{x} = f(x