Differential Equations: An Introduction

# Differential Equations: An Introduction ## 1. Differential Equations - An equation involving differentials or differential coefficients or derivatives is called a differential equation. - **Examples:** - $ \frac{dy}{dx} = x + 3 $ - $ \frac{d^2y}{dx^2} + 3\frac{dy}{dx} + 2y = 0 $ -...

# Differential Equations: An Introduction ## 1. Differential Equations - An equation involving differentials or differential coefficients or derivatives is called a differential equation. - **Examples:** - $ \frac{dy}{dx} = x + 3 $ - $ \frac{d^2y}{dx^2} + 3\frac{dy}{dx} + 2y = 0 $ - $ y + xy\frac{dz}{dy} = k\left(1 + \frac{dy}{dx}\right)^2 $ - $ p= \frac{d^2y/dx^2}{1+dy/dx}$ - $ y \frac{dy}{dx} = x\frac{dy}{dx} + a $ - All of these are **differential equations**, which can be defined as those equations in which all the differential coefficients have reference to a single independent variable. - **Partial differential equations**, which are beyond the scope of this document, contain differential coefficients referring to more than one independent variable. ## 2. Order and Degree of Differential Equation - The **order** of a differential equation is the order of the highest derivative which appears in it. - **Example:** The order of the equation $ 1 + (\frac{dy}{dx})^{3/2} = p\frac{d^2 y}{dx^2}$ **is two**, as the highest derivative $ \frac{d^2 y}{dx^2} $ appears in the equation. - The **degree** of a differential equation is the degree of the highest derivative that appears in it after the equation has been made clear of radicals and fractions. ## 1.2.1 Solutions of Differential Equations - A solution or integral of a differential equation is a relation that exists between the dependent and independent variables and at the same time the derivatives obtained there from satisfy the given equation. - **Example:** $y = sin x + c$ is a solution of the differential equation $ \frac{dy}{dx} = cos x$. ## 1.2.2 Kinds of Solutions - **General Solution or the Complete Primitive**: The solution of a differential equation which contains a number of arbitrary constants equal to the order of the differential equation is called the general solution or the complete primitive. - **Example**: $y = c_{1} cos x + c_{2} sinx$ is a general solution of the differential equation $ \frac{d^2y}{dx^2} + y = 0 $ - **Particular Solution**: If particular values are given to the arbitrary constants in the general solution, then the solution so obtained is called the particular solution. - **Example**: For the complete primitive $ y = A cos (x + B)$, if the constants $A$ and $B$ are assigned to the particular values $1$ and $0$, the particular solution is $y = cos x$ - **Singular Solution**: The solution which cannot be derived from the complete primitive or general solution by assigning particular values to the arbitrary constants are called the Singular Solutions. ## Examples **1.** Find the differential equation of the family of the curves $y = Ae^{2x} + Be^{-2x}$, for different values of A and B. - **Solution:** - We have $y = Ae^{2x} + Be^{-2x}$ - Differentiating both sides with respect to $x$, we get $ \frac{dy}{dx} = 2Ae^{2x}-2Be^{-2x}$ - Differentiating again, we get $ \frac{d^2y}{dx^2} = 4Ae^{2x} + 4Be^{-2x} = 4(Ae^{2x} + Be^{-2x}) = 4y$. - **Therefore, the required differential equation is $ \frac{d^2y}{dx^2} = 4y $** **2.** Find the differential equation of the system of curves $ y=ax^2 + b cos nx + c$, where a, b, c are arbitrary constants. - **Solution:** - We have $ y=ax^2+b cos nx+c$. - Differentiating, we get $ \frac{dy}{dx} = 2ax - bn sin nx$ - Differentiating again, we get $ \frac{d^2y}{dx^2} = 2a - bn^2 cos nx$ - Differentiating once more, we get $ \frac{d^3y}{dx^3} = bn^3 sin nx$ - Eliminating a between (ii) and (iii), we get $ x^{2}y\frac{d^2y}{dx^2} - \frac{dy}{dx} = bn(cos nx - sin nx)$ - Eliminating b between (iv) and (v) by dividing, we get $ \frac{d^3y}{dx^3} / \frac{x^2y\frac{d^2y}{dx^2}-\frac{dy}{dx}}{x^2y}$ = $ \frac{-x^2 sin nx}{cos nx -sin nx}$ - **This is the required differential equation.** Note that here three constants have been eliminated and we have got a differential equation of the third order. ## Exercises 1. Find the differential equation from $x^2 - y² + 2\lambda xy = 1$, where λ is a parameter. 2. Determine a differential equation from the equation $ax^2 + by^2 = 1$, where a and b are parameters. 3. From the differential equation of all circles of radius r. 4. Form the differential equation of all circles of radius r. 5. Show that v = A/r + B is a solution of $ \frac{d^2v}{dr^2}$ + $ \frac{2}{r}\frac{dv}{dr} = 0$. 6. Find the differential equation corresponding to y = ae^2x + be^-3x + ce^x. ## 2. Differential Equations of First Order and First Degree ## 2.1 Equations of the First Order and First Degree - The general form of an ordinary differential equation of the first order and first degree can be expressed in the form $M + N \frac{dy}{dx} = 0$, where M and N are functions of x and y, or are constants. - The equation can also be expressed as: - $\frac{dy}{dx} = f(x, y)$ - $(x, y) dy = f(x, y) dx$ - $-f(x, y) dx + (x, y) dy = 0$ - $Mdx + Ndy = 0$ ## 2.2 Equations in Which the Variables Are Separable - Variables are separable if a differential equation as can be written as $ f(x) dx + (y) dy = 0$ - To solve such equations, integrate both sides with respect to the respective variables: - $\int f(x)dx + \int (y) dy = c$, where c is an arbitrary constant. ## Working Rule: 1. Arrange all the terms containing x together with dx on one side and all the terms of y with dy on the other side. 2. Integrate each term separately with respect to either x or y as associated with dx or dy. 3. Add an arbitrary constant on one side. ## Examples: **1.** Solve: $ \frac{dy}{dx} - \frac{x}{y} = \frac{y^2}{x^2} $ - **Solution:** - The equation can be written as: $ (y-ay²) = (x+a) \frac{dy}{dx} $ - Rewriting the equation to separate the variables, we get: $ \frac{1}{y(1 -ay)} \frac{dy}{dx} = (x+a) $ - Integrating both sides, we get: $ log y -log(1-ay) = log(x+a) + log c $ - Simplifying, we get $ log \frac{y}{1 -ay} = logc(x+a) $ - Final solution: $y = c(1 - ay)(x+a)$ **2.** Solve: $ 3e^x tan y + (1 - e^x)sec^2 y \frac{dy}{dx} = 0$ - **Solution:** - Substituting $tan y= v$, we get $ 3e^xv + (1 -e^x) \frac{dv}{dx}= 0$ - Integrating both sides, we get: $3 log (e^x - 1) = log v + log c = log vc$ - Simplifying the equation, we get: $ (e^x-1)^3 = cv$ or $ (e^x-1)^3 = c tan y$ ## Exercise: Solve the following differential equations: 1. $ \frac{dy}{dx} = \frac{1+y^2}{1+x^2}$ 2. $ \frac{dy}{dx} -x+2y$ 3. $ \frac{dy}{dx} = \frac{e^y + x^2}{e^y +e^x}$ 4. $ (x+2) \frac{dy}{dx} = x\frac{dy}{dx} + y$ 5. $ \frac{dy}{dx} = 2y(2 log x+1)$ 6. $ (x^2 + x) dx + (x² + y)dy = 0$ ## 1.3 Change of Variables - Sometimes a substitution can be used to reduce a given differential equation to a form in which the variables are separable. ## Examples: **1.** Solve: $ (x + y)² \frac{dy}{dx} = a^2$ - **Solution:** - Put $x + y = v$, then $\frac{dy}{dx} = dv - 1$ - Substituting these values, we get: $v²(-1) = a^2$ - Rearranging, we get $v² dv = (v^2 + a^2)dx$ - Simplifying, we get $dx = \frac{v²}{(v^2 + a^2)}dv$ - Integrating both sides, we get $x = v-atan^{-1} \frac{v}{a} + c$ - Replacing v with x + y, we get $x = x+y-atan^{-1} \frac{x+y}{a} + c$ - **Solution:** $atan^{-1} \frac{x+y}{a} = x+ y$ **2.** Solve: $ x\frac{dy}{dx}-y\frac{dx}{dy} = \frac{x² +y²}{a^2 -x^2 - y²}$ - **Solution** - Substitute $ x = r cos 0$, $ y = r sin 0$ - We know that $ x^2 + y^2 = r^2$ - By differentiation $ dx = dr cos 0 - r sin 0 d0$ $ dy = dr sin 0 + r cos 0 d0$ - Replacing these substitutions, we get $ \frac{r cos 0}{r sin 0} * \frac{dr}{d0} * \frac{d0}{dr} = \frac{r^2}{a² - r²}$ - Simplifying, we get $ cot 0 (cos 0 - r sin 0) d0 = \frac{r}{a^2 - r^2} (sin 0 + r cos 0) d0$ - **The resulting equation is now separable and can be solved using integration.** ## Exercise: Solve the following differential equations: 1. $(x-y)^2 \frac{dy}{dx} = a^2$ 2. $\frac{dy}{dx} = sin(x+y)+ cos(x+y)$ 3. $sin(\frac{dy}{dx}) = x+y$ 4. $x \frac{dy}{dx} - y \frac{dx}{dy} = (x^2 + y^2) dx$ 5. $\frac{dy}{dx} - xtan(y-x) =1$ 6. $\frac{x+y-b}{x+y-b} dy = \frac{x+y + a}{x+y+b} dx$ 7. $sin y \frac{dy}{dx} = cos y (1- x cos y)$ 8. $sec^2 y \frac{dy}{dx} + 2x tan y = x^3$ 9. $\frac{dy}{dx} = e^{x-y} (e^x-e^x)$ ## 2.4 Homogeneous Equations - Homogeneous Differential Equations are of the type: $\frac{dy}{dx} = \frac{f_1(x, y)}{f_2(x, y)}$ where $f_1(x, y)$ and $f_2(x, y)$ are homogeneous functions of x and y of the same degree. - **To solve these equations:** 1. Put y = vx (or x = vy) 2. Substitute this into the given equation in terms of v and x (or v and y) 3. Apply the method of separation of variables 4. After integration, replace v by y/x (or x/y) ## Examples: **1.** Solve: $x^2y dx - (x^3 + y³) dy = 0$ - **Solution:** - This is a homogeneous differential equation. To solve it, substitute $y = vx$. - Then the equation becomes $x²y = (x^3 + v^3x^3)$ - Simplifying, we get $ v + x\frac{dv}{dx}= \frac{v}{1+v²}$ - Separating the variables, we get $ \frac{dv}{1 + v³} = \frac{1}{x}+ \frac{-v}{1+v³} dx$ - Integrating both sides, we get $ \frac{1}{3} (-\frac{1}{v} + \frac{1}{2} log(1 + v + v²) + \frac{1}{2} log(1 - v + v²)) = log x + c$ ## Exercise: Solve the following differential equations: 1. $(x^2 - y²) dx + 2xy dy = 0$ 2. $2xy dx + (y² - x²) dy = 0$ 3. $(x^2 + y²) dx + 2xy dy = 0$ 4. $(x² + xy) dy = (x² + y²) dx$ 5. $x(x - y) dy + y² dx = 0$ 6. $cxy = e^{-x/y}$ 7. $\{x\sqrt{x^2 + y^2} - y^2\}dx + xy dy = 0$ 8. $y² + x^2 \frac{dy}{dx} = xy\frac{dy}{dx}$ 9. $x^2 dy + y (x + y) dx = 0$ 10. $x²\frac{dy}{dx} = y(x+y)$ ## 2.5 Equations Reducible to a Homogeneous Form - All the differential equations of the type: $\frac{dy}{dx} = \frac{ax+by+c}{Ax+By+C}$ can be reduced to the homogeneous form using the following substitution: $ X = x - h$ and $Y = y - k$ ## Working Rule: 1. Put $x = X + h, y = Y + k$ in the given differential equation. 2. Equate the constant terms (independent of $X$ and $Y$ ) in the numerator and denominator to zero. 3. Solve these resulting equations for $h$ and $k$. 4. Solve the resultant homogeneous equation in X and Y by the method of 2.4. 5. Replace X by x - h and Y by y - k 6. Substitute the values of h and k. ## Examples: **1.** Solve: $(2x + y + 3) dx - (2y + x + 1) dy = 0$ - **Solution:** - We have a = 2, b = 1, A = 2, B = 1, c = 3 and C = 1 - Substituting $x = X + h, y = Y+ k$ in the equation and equating the constant terms to $0$, we get: $ 2h+ k + 3 = 0$ and $h + 2k + 1 = 0$ - Solving these equations, we get $h = -5/3$ and $k = -1/3$ - Substituting these values, the equation becomes $\frac{dY}{dX} = \frac{2X + Y}{2Y + X}$ - Now, substitute $Y = vX$, we get $\frac{dv}{dX} = \frac{2 + v}{2v + 1}$ - Integrating both sides, we get $log(1-v) - log(1+v) = 4log X - log c$ - Simplifying, we get $(X + Y)(X - Y)^3 = c$ - And replacing X by $ x + 5/3$ and $Y$ by $y + 1/3$, we get $ (x + y + 1)(x - y + 2)^3 = c$ **2.** Solve $ \frac{dy}{dx}$ = $\frac{x - y + 3}{2x - 2y + 5}$ - **Solution:** - Let $x - y = v$. - Making the substitution, we get $ \frac{dv}{dx} = \frac{v + 3}{2v + 5}$ - Multiplying both sides by $(2v + 5)$ and integrating, we get $2v + log(2v+5) = x+ c$ - We know that $v = x-y$, therefore $2(x - y) + log (x - y + 5) = x + c$ - **The solution is:** $x - 2y + log(x-y+5) = c$ ## 2.6 Linear Differential Equations - A first order differential equation is linear if it can be expressed as: $\frac{dy}{dx} + Py = Q$, where P and Q are functions of x only, or constants. - We can also rewrite the equation to be expressed as $\frac{dy}{dx} + Px = Q$, where P and Q are functions of y only, or constants. This is also a linear differential equation. ## 2.6.1 Solution of Linear Equations - The given equation$\frac{dy}{dx} + Py = Q$. - The **integrating factor** of this equation is $e^{\int Pdx}$. ## Working Rule: 1. Find the integrating factor $e^{\int Pdx}$, which is represented as the I.F. 2. Multiply the differential equation by I.F. ($e^{\int Pdx}$), and integrate both sides using integration by parts - taking $e^{\int Pdx}$ as the first function. 3. If x is the independent variable, the I.F. is $e^{\int Pdy}$. ## Examples: **1.** Solve: $(1+x^2) \frac{dy}{dx}$ + $2xy = y$. - **Solution**: - We can rewrite the equation as: $ \frac{dy}{dx} + \frac{2x}{1+x^2}y = \frac{y}{1+x^2}$ - The integrating factor is $e^{\int \frac{2x}{1+x^2} dx} = e^{log(1+x^2)} = (1 + x^2)$ - Multiplying the equation by I.F and integrating, we get $ y(1+x^2) = \frac{1}{2} \int \frac{y}{1+x^2} dx + c$ - The solution is: $y(1+x^2) = \frac{1}{2} tan^{-1} x + c$ **2.** Solve: $(1 + y²) \frac{dy}{dx} + (x - e^{tan^{-1}y}) y = 0$. - **Solution**: - The equation can be written as $(1 + y²) \frac{dy}{dx} + xy = e^{tan^{-1}}y$. - The integrating factor is $e^{\int \frac{x}{1+y^2} dx} = e^{\frac{1}{2} tan^{-1}y}$. - Multiplying the equation by I.F. and integrating, we get $ xe^{tan^{-1}y} = \int \frac{e^{tan^{-1}y}}{1+y^2} * y * e^{\frac{1}{2} tan^{-1}y} dy + c$ - Substituting $t$ for $tan^{-1}y$, we get $xe^{tan^{-1}y} = \int \frac{e^{2t}}{2} dt + c$ - Integrating, we get $xe^{tan^{-1}y} = \frac{1}{2}e^{2 tan^{-1}y} + c$ - **The solution is:** $xe^{tan^{-1}y} - \frac{1}{2}e^{2 tan^{-1}y} = c$ **3.** Solve: $\sqrt{a² + x²} \frac{dy}{dx} + y = \sqrt{a² + x²} - x$ - **Solution:** - We can rewrite the equation as $\frac{dy}{dx} + \frac{y}{\sqrt{a² + x²}} = 1 - \frac{x}{\sqrt{a² + x²}}$ - The integrating factor is $ e^{\int \frac{1}{\sqrt{a² + x²}} dx} = e^{log (\sqrt{a²+x²} + x)} = (\sqrt{a²+x²} + x)$ - Multiplying the equation by the integrating factor, we get $ (\sqrt{a²+x²} + x)\frac{dy}{dx} + \frac{y(\sqrt{a²+x²} + x)}{\sqrt{a²+x²}}= (\sqrt{a²+x²} + x) - x) $ - Simplifying the equation, we get $ (\sqrt{a²+x²} + x)\frac{dy}{dx} + y = \sqrt{a²+x²}$ - Integrating both sides, we get: $y(\sqrt{a²+x²} + x) = \int \sqrt{a²+x²} dx + c $ - **The solution is:** $y (\sqrt{a²+x²} + x) = \frac{1}{2} [x\sqrt{a²+x²} + a² sinh^{-1}\frac{x}{a} ]+ c$ ## Exercise: Solve the following differential equations: 1. $\frac{dy}{dx} + \frac{2x}{1+x^2} y = \frac{1}{(1+x^2)^2}$ 2. $(x + y +1)\frac{dy}{dx} = 1$ 3. $\frac{dy}{dx} + y tan x = sec x$ 4. $\frac{dy}{dx} + y tan x - sec x = 0$ 5. $\frac{dy}{dx} - y tan x = 2sin x$ 6. $\frac{dy}{dx} + y cot x = x^2$ 7. $\frac{dy}{dx} + 2y = x²$ if $y = 1$ when $x = 1$ 8. $(1+x^2) \frac{dy}{dx} + y = tan x$ 9. (1+y²) (x-e^{-tany}) dy = 0 10. (1 + y²) dx = (tan^{-1} y - x) dy 11. x log x + y = 2 log x 12. (2x - 10 y³) $\frac{dy}{dx}$ + y = 0 13. x $\frac{dy}{dx}$ - y = 2x² cosec 2x 14. $\frac{dy}{dx} + \frac{y}{x + \sqrt{1-x²}}$ = $(1 + \frac{1}{1-x²})²$ 15. $cos^3x \frac{dy}{dx} + y cos x = sin x$ ## 2.7 Differential Equations Reducible to the Linear Form - A Bernoulli's equation is a differential equation of the form: $\frac{dy}{dx}$ + Py = Qy^n$ where P and Q are functions of x only. ## Solving Bernoulli's Equations: - Substitute $v = y^{1-n}$, then differentiate to get $dv = (1 – n) y^{-n} dy$ - Substituting into the original equation, we get $ (1 - n) y^{-n} \frac{dy}{dx}$ + $(1 - n) P y^{1-n}$ = $(1 – n) Q$. - **This is a linear equation.** We can solve it using the integrating factor = $ e^{\int (1-n) Pdx}$ ## Working Rule: 1. Write the equation in the form: $ \frac{dy}{dx}$ + Py = Qy^n$ 2. Divide both sides by y^n to bring it to the form as in § 2.7. 3. Substitute $v = y^{1-n}$ and change the equation to the form $ \frac{dv}{dx} = (1 – n) Q - (1 – n) P v$ 4. Then apply § 2.7 to solve. ## Examples: **1.** Solve: $\frac{dy}{dx} - \frac{tan y}{1+x} = (1 + x)e^x sec y$ - **Solution**: - Dividing both sides by $sec y$ we get: $ \frac{dy}{dx} - \frac{sin y}{cos y} = (1 + x) e ^x$ - Substituting $v = sin y$, we get $ \frac{dv}{dx} = (1 + x)e^x cos y$ - Substituting the value of $cos y = \sqrt{1 - sin² y}$, we get $ \frac{dv}{dx} = \frac{(1 + x)e^x}{\sqrt{1 - v²}}$ - The integrating factor is $ e^{\int \frac{(1 + x)}{1 + x^2} dx} = e^{\frac {1}{2} log(1 + x²)} = \sqrt{1 + x^2}$ - Multiplying both sides by the I.F., we get $ v\sqrt{1 + x²}$ = $\int (1 + x) e^x dx + c$ - Integrating both sides, we get $v\sqrt{1 + x²} = xe^x + c$ - Replacing $v$ by $sin y$ we get: $sin y\sqrt{1+x^2} = xe^x + c$ **2.** Solve: $y \frac{dy}{dx} - y = a² + \frac {dy}{dx}$ - **Solution:** - We have a Bernoulli’s equation, here n = 2. - Substituting v = y^-1, we get $\frac{dv}{dx} = -v²\frac{dy}{dx}$ - Replacing in the original equation, we get $(x+a)\frac{dv}{dx} + v^2 = -a$ - Dividing both sides by $-v^2 (x + a)$, we get $\frac{1}{x + a} \frac{dv}{dx} - \frac{1}{v^2} = \frac{a}{x + a}$ - The I.F. is $e^{\int \frac{1}{x + a} dx} = e^{log (x+a) } = (x + a)$ - Multiplying the equation by I.F., we get $ v(x + a) = \int \frac{a}{x+a} dx + c$ - Integrating both sides, we get $v(x + a) = a log(x + a) + c$ - Replacing $v = \frac{1}{y}$, we get $\frac{x+a}{y} = a log(x + a) +c$ - **The solution is:** $(x + a)y = alog(x+a) + c)$ **3.** Solve: $ \frac{dz}{dx}. z + \frac{z}{x}$ = $(z log z)^2$ - **Solution:** - We can rewrite this equation in the form $ \frac{dz}{dx}. z + \frac{1}{x}z = z^2 log^2z$ - Dividing both sides by $z²$, we get: $ \frac{1}{z} \frac{dz}{dx} + \frac{1}{xz} = log^2 z$ - Substituting $v= log z$ and differentiating, we get $ \frac{dv}{dx} = \frac{1}{z}. \frac{dz}{dx}$, and $\frac{dv}{dx}$ = $\frac{1}{xz}dx$ - Replacing in the original equation, we get $ \frac{dv}{dx} + \frac{1}{x}v = v^2$. - The integrating factor is $e^{\int \frac{1}{x} dx} = e^{log x} = x$ - Multiplying the equation by the I.F. and integrating both sides, we get $ x log z = \frac{1}{2} (log z)^2 + C$ - **The solution is:** $x log z = \frac{(log z)^2}{2} + C$ ## Exercise: Solve the following differential equations: 1. $ 2\frac{dy}{dx} - \frac{y}{x} = y²$ 2. $x\frac{dy}{dx} - \frac{y}{x} = x^2$ 3. $\frac{dy}{dx} + 2ytan x = y^2$ 4. $\frac{dy}{dx} + y co^t x = y^ 2 sin^2x$ 5. $ d(x²y + xy) = 1$ 6. $(x^3y^2 + xy) dx$ = $dy$ 7. $\frac{dy}{dx} + \frac{y}{x} = y^2e^{\frac{x^2}{2} } sin x$ 8. $x \frac{dy}{dx} + y log y = tan y sin y$ ## 2.8 Exact Differential Equation - An exact differential equation is a differential equation which can be derived from its primitive by direct differentiation without any further transformation (like elimination or reduction). - **Example:** The differential equation $(ax + hy + g) dx + (hx + by + f) dy = 0$ is exact because it is derived from its primitive $ \frac{ax²}{2} + \frac{by²}{2} + hxy + gx + fy + c = 0$ by direct differentiation. ## 2.8.1 Theorem - **The necessary and sufficient condition for the ordinary differential equation Mdx + Ndy = 0 to be exact is:** $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ ## 2.8.2 Finding Integrating Factors: - When an equation Mdx + Ndy = 0 is not exact but is homogeneous, then $\frac{1}{Mx+Ny}$ is an integrating factor. - When an equation Mdx + Ndy = 0 is not exact but is written in the form f(xy) ydx + f2(xy) xdy = 0, then $\frac{1}{Mx - Ny}$ is an integrating factor provided Mx - Ny ≠ 0. ## Examples: **1.** Solve: $(x^2 - 2xy^2)dx - (x^3 - 3x^2y) dy = 0$ - **Solution**: - M = x² - 2xy², N = 3x²y - x³ - We have $ \frac{\partial M}{\partial y} = -4y$, and $ \frac{\partial N}{\partial x} = 6xy - 3x²$ - The equation is not exact, because $ \frac{\partial M}{\partial y}$ ≠ $ \frac{\partial N}{\partial x}$ - Since M and N are homogeneous functions, we find the integrating factor: $\frac{1}{Mx+Ny}= \frac{1}{x^2y²}$ - Multiply both sides by the integrating factor: $ (\frac{1}{2} + \frac{1}{x^{2}y^{2}}) dx - (\frac{3}{2}x - \frac{1}{x^{2}y^{2}}) dy =0$ - We can prove that $ \frac{\partial (\frac{1}{2} + \frac{1}{x^{2}y^{2}})}{\partial y} = \frac{\partial (\frac{3}{2}x - \frac{1}{x^{2}y^{2}})}{\partial x}$ - The equation is exact and its solution is: $ \int (\frac{1}{2} + \frac{1}{x^{2} y^{2}}) dx + \int(- \frac{1}{x^{2} y^{2}}) dy = c$ - The solution is: $\frac{1}{2}x - \frac{1}{xy} = c$ **2.** Solve: $(x^3y³ + x^2y² + xy + 1) ydx + (x^3y³ - x²y² - xy + 1) xdy = 0$ - **Solution:** - We have M = $xy(x^3y³ + x^2y² + xy + 1)$ N = $ xy(x^3y³ - x^2y² - xy + 1)$ - To solve this problem, the integrating factor is: $\frac{1}{Mx - Ny} = \frac{1}{2x^2y^2(xy +1)} $ - Multiplying by the integrating factor, we get: $\frac{1}{2}dx + \frac{(xy +1)(x^2y^2 - 2xy + 1)}{2x^2y^2(xy + 1)} dy = 0$ - Simplifying the