First Order Differential Equations

If the variables of a first-order differential equation can be grouped on either side of the equation, a solution can be found by integrating both sides. This method is called separation of variables. (Activity 7.2 is omitted as it describes a process rather than providing examples.) A separable equation is a first-order differential equation where $f(x, y)$ can be expressed as a product of a function of $x$ only and a function of $y$ only. That is, $\frac{dy}{dx} = f(x, y) = g(x)h(y)$, which can be separated as $\frac{dy}{h(y)} = g(x)dx$.

Example 1

Write each of the following first-order differential equations in general form and then separate the variables: (a) $dy - x^2e^{5x}y^4e^{-3y}dx = 0$ (b) $\sin x \sin y dx + \cos x \cos y dy = 0$ (c) $(1 + x)dy - ydx = 0$ (d) $(x^2 + 1)dy + (y^2 + 1)dx = 0$

Solution: (a) $dy = x^2e^{5x}y^4e^{-3y}dx$ $\frac{dy}{dx} = x^2e^{5x}y^4e^{-3y}$ $\frac{e^{3y}}{y^4}dy = x^2e^{5x}dx$ (b) $\cos x \cos y dy = -\sin x \sin y dx$ $\frac{dy}{dx} = -\frac{\sin x \sin y}{\cos x \cos y}$ $\frac{\cos y}{\sin y}dy = -\frac{\sin x}{\cos x}dx$ $\cot y dy = -\tan x dx$ (c) $(1 + x)dy = ydx$ $\frac{dy}{dx} = \frac{y}{1 + x}$ $\frac{dy}{y} = \frac{dx}{1 + x}$ (d) $(x^2 + 1)dy = -(y^2 + 1)dx$ $\frac{dy}{dx} = -\frac{y^2 + 1}{x^2 + 1}$ $\frac{dy}{y^2 + 1} = -\frac{dx}{x^2 + 1}$

Example 2

Separate the variables in each of the following differential equations: (a) $2\sin y \cos y dy - \sin 2y \cos x dy = 4e^{2x}e^3 \sin x dx$ (b) $(x^4 + \theta^4 + 1 + x^4)dx = (\theta^2)(x^2 + 1)d\theta$

Solution: (a) $2\sin y \cos y dy - \sin 2y \cos x dy = 4e^{2x}e^3 \sin x dx$ $\sin 2y dy - \sin 2y \cos x dy = 4e^{2x+3} \sin x dx$ $\sin 2y(1-\cos x) dy = 4e^{2x+3} \sin x dx$ $\frac{\sin 2y dy}{1} = \frac{4e^{2x+3} \sin x dx}{1-\cos x}$ $\sin 2y dy = \frac{4e^{2x+3} \sin x}{1-\cos x}dx$ (b) $(2x^4 + \theta^4 + 1)dx = \theta^2(x^2 + 1)d\theta$ $\frac{2x^4 + \theta^4 + 1}{x^2 + 1}dx = \theta^2 d\theta$

If the given equation is of the form $\frac{dy}{dx} = f(x, y)$, where $f(x, y)$ can be written as $g(x)h(y)$, then separating the variables gives $\frac{dy}{h(y)} = g(x)dx$. Integrating both sides gives the solution: $\int \frac{dy}{h(y)} = \int g(x)dx$

An initial-value problem consists of a differential equation and one or more supplementary conditions that are all related to the same value of the independent variable (often at an initial time, hence the name). A boundary-value problem has supplementary conditions related to different values of the independent variable.

Examples: (a) $\frac{d^2y}{dx^2} + y = 0; y(1) = 3, y'(1) = -4$ is an initial-value problem. (b) $\frac{d^2y}{dx^2} + y = 0; y(0) = 0, y(\pi) = 1$ is a boundary-value problem.

Example 1

Solve the differential equation $\frac{dy}{dx} = \frac{y^2(x + 3)}{x^2(y + 1)}$.

Solution: Separating variables: $\frac{y + 1}{y^2}dy = \frac{x + 3}{x^2}dx$ $\left(\frac{1}{y} + \frac{1}{y^2}\right)dy = \left(\frac{1}{x} + \frac{3}{x^2}\right)dx$ Integrating both sides: $\int \left(\frac{1}{y} + \frac{1}{y^2}\right)dy = \int \left(\frac{1}{x} + \frac{3}{x^2}\right)dx$ $\ln|y| - \frac{1}{y} = \ln|x| - \frac{3}{x} + C$ This is the general solution. It's difficult to express $y$ explicitly in terms of $x$ in this case.

Example 2

Solve the differential equation $\frac{dy}{dx} = \frac{x - 5}{y^2}$, for $y(0) = 3$, expressing $y$ in terms of $x$.

Solution: Separating variables: $y^2dy = (x - 5)dx$ Integrating both sides: $\int y^2dy = \int (x - 5)dx$ $\frac{y^3}{3} = \frac{x^2}{2} - 5x + C$ Applying the initial condition $y(0) = 3$: $\frac{3^3}{3} = \frac{0^2}{2} - 5(0) + C$ $9 = C$ So, $\frac{y^3}{3} = \frac{x^2}{2} - 5x + 9$ $y^3 = \frac{3x^2}{2} - 15x + 27$ $y = \sqrt[3]{\frac{3}{2}x^2 - 15x + 27}$

Example 3

Solve the initial-value problem $(2\cos x + \cos y)dy + (\sin x + \sin y)dx = 0$, $y(0) = \frac{\pi}{2}$.

Solution: This equation is not directly separable. However, it can be rearranged as: $2\cos x dx + \sin x dx + \cos y dy + \sin y dy = 0$ Integrating each term: $2\int\cos x dx + \int\sin x dx + \int\cos y dy + \int\sin y dy = \int 0$ $2\sin x - \cos x + \sin y - \cos y = C$ Applying the initial condition $y(0) = \pi/2$: $2\sin(0) - \cos(0) + \sin(\pi/2) - \cos(\pi/2) = C$ $0 - 1 + 1 - 0 = C$ $C = 0$ Thus, the solution is: $2\sin x - \cos x + \sin y - \cos y = 0$ $\sin y = \cos y - 2\sin x + \cos x$

Some first-order differential equations are not separable in their original form but can be made separable by a change of variables. A first-order differential equation $M(x, y)dx + N(x, y)dy = 0$ is said to be homogeneous if it can be written in the form $\frac{dy}{dx} = f\left(\frac{y}{x}\right)$, where the functions $M(x, y)$ and $N(x, y)$ are homogeneous of the same degree. In this case, $\frac{dy}{dx}$ is isolated on one side, while the other side has an expression where $y$ and $x$ appear in the combination $\frac{y}{x}$. For instance, $\frac{dy}{dx} = 1 + \frac{y}{x}$ is a homogeneous first-order differential equation. A homogeneous differential equation can be transformed into a separable equation by the following change of variables: From $\frac{dy}{dx} = f\left(\frac{y}{x}\right)$, let $u = \frac{y}{x}$. Replacing $\frac{y}{x}$ with $u$ in the differential equation gives: $\frac{dy}{dx} = f(u)$ (7.1) Since $u = \frac{y}{x}$, it follows that $y = ux$. Differentiating $y = ux$ with respect to $x$ gives: $\frac{dy}{dx} = u + x\frac{du}{dx}$ Substituting $\frac{dy}{dx}$ into equation (7.1) gives: $u + x\frac{du}{dx} = f(u)$ $x\frac{du}{dx} = f(u) - u$ (7.2) Equation (7.2) is now a separable differential equation. Rearranging the variables gives: $\frac{du}{f(u) - u} = \frac{dx}{x}$ (7.3) Equation (7.3) can now be solved by integrating both sides.

Example 1

Determine whether or not each of the following differential equations is homogeneous: (a) $\frac{dy}{dx} = \frac{y + x}{x}$ (b) $x^2\frac{dy}{dx} = y^2 + x^2$ (c) $\frac{dy}{dx} = \frac{y^2}{x}$ (d) $\frac{dy}{dx} = \frac{x^2 + y}{x^3}$

Solution: (a) $\frac{dy}{dx} = \frac{y + x}{x} = \frac{y}{x} + 1$. This is in the form $\frac{dy}{dx} = f\left(\frac{y}{x}\right)$. Therefore, it is homogeneous. (b) $\frac{dy}{dx} = \frac{y^2 + x^2}{x^2} = \left(\frac{y}{x}\right)^2 + 1$. This is in the form $\frac{dy}{dx} = f\left(\frac{y}{x}\right)$. Therefore, it is homogeneous. (c) $\frac{dy}{dx} = \frac{y^2}{x} = \frac{y}{x} \cdot y$. This cannot be written solely as a function of $\frac{y}{x}$. Therefore, it is not homogeneous. (d) $\frac{dy}{dx} = \frac{x^2 + y}{x^3} = \frac{1}{x} + \frac{y}{x^3}$. This cannot be written solely as a function of $\frac{y}{x}$. Therefore, it is not homogeneous.

Example 2

Solve each of the following differential equations: (a) $\frac{dy}{dx} = 1 + \frac{y}{x}$ (b) $x^2dy = (y^2 + xy + x^2)dx; y(1) = 0$

Solution: (a) Let $u = \frac{y}{x}$, so $y = ux$. Then $\frac{dy}{dx} = u + x\frac{du}{dx}$. Substituting into the given equation: $u + x\frac{du}{dx} = 1 + u$ $x\frac{du}{dx} = 1$ $du = \frac{dx}{x}$ $\int du = \int \frac{dx}{x}$ $u = \ln|x| + C$ Substituting back $u = \frac{y}{x}$: $\frac{y}{x} = \ln|x| + C$ $y = x\ln|x| + Cx$ (b) $\frac{dy}{dx} = \frac{y^2 + xy + x^2}{x^2} = \left(\frac{y}{x}\right)^2 + \frac{y}{x} + 1$ Let $u = \frac{y}{x}$, so $y = ux$ and $\frac{dy}{dx} = u + x\frac{du}{dx}$. Substituting: $u + x\frac{du}{dx} = u^2 + u + 1$ $x\frac{du}{dx} = u^2 + 1$ $\frac{du}{u^2 + 1} = \frac{dx}{x}$ $\int \frac{du}{u^2 + 1} = \int \frac{dx}{x}$ $\arctan(u) = \ln|x| + C$ $\arctan\left(\frac{y}{x}\right) = \ln|x| + C$ Using the initial condition $y(1) = 0$: $\arctan(0) = \ln(1) + C$ $0 = 0 + C$ $C = 0$ $\arctan\left(\frac{y}{x}\right) = \ln|x|$ $y = x\tan(\ln|x|)$

Example 3

Find the solution in each of the following initial-value problems: (a) $x\frac{dy}{dx} = \frac{y^2}{x} + y; y(1) = 1$ (b) $(x + 2y + 3)dy + (2x + y + 6)dx = 0; y(-1) = 1$ (c) $2xyy' - y^2 + x^2 = 0; y(-2) = 3$

Solution: (a) $\frac{dy}{dx} = \frac{y^2}{x^2} + \frac{y}{x}$. Let $u = \frac{y}{x}$, so $y = ux$ and $\frac{dy}{dx} = u + x\frac{du}{dx}$. $u + x\frac{du}{dx} = u^2 + u$ $x\frac{du}{dx} = u^2$ $\frac{du}{u^2} = \frac{dx}{x}$ $\int \frac{du}{u^2} = \int \frac{dx}{x}$ $-\frac{1}{u} = \ln|x| + C$ $-\frac{x}{y} = \ln|x| + C$ Using $y(1) = 1$: $-1 = \ln(1) + C \Rightarrow C = -1$ $-\frac{x}{y} = \ln|x| - 1$ $y = \frac{-x}{\ln|x| - 1} = \frac{x}{1 - \ln|x|}$ (b) This equation is not homogeneous in its current form. It requires a different substitution. Let $u = x+2y+3$ and $v=2x+y+6$. Then $2y=u-x-3$ and $y=v-2x-6$. Substituting into the original equation and solving for $dy/dx$ doesn't lead to a simple homogeneous form. This problem is of the form $(a_1x+b_1y+c_1)dx+(a_2x+b_2y+c_2)dy=0$. The method for this type of problem is not covered in the previous text. Thus it is not solved here. (c) $\frac{dy}{dx} = \frac{y^2 - x^2}{2xy} = \frac{1}{2}\left(\frac{y}{x} - \frac{x}{y}\right)$. Let $u = \frac{y}{x}$, so $y = ux$ and $\frac{dy}{dx} = u + x\frac{du}{dx}$. $u + x\frac{du}{dx} = \frac{1}{2}\left(u - \frac{1}{u}\right)$ $x\frac{du}{dx} = \frac{u}{2} - \frac{1}{2u} - u = -\frac{u}{2} - \frac{1}{2u} = -\frac{u^2 + 1}{2u}$ $\frac{2u}{u^2 + 1}du = -\frac{dx}{x}$ $\int \frac{2u}{u^2 + 1}du = -\int \frac{dx}{x}$ $\ln(u^2 + 1) = -\ln|x| + C$ $\ln\left(\frac{y^2}{x^2} + 1\right) = -\ln|x| + C$ $\ln\left(\frac{y^2 + x^2}{x^2}\right) = \ln\left|\frac{A}{x}\right|$ (where $C = \ln|A|$) $\frac{y^2 + x^2}{x^2} = \frac{A}{x}$ $y^2 + x^2 = Ax$ Using $y(-2) = 3$: $3^2 + (-2)^2 = A(-2)$ $9 + 4 = -2A$ $13 = -2A$ $A = -\frac{13}{2}$ Therefore, $y^2 + x^2 = -\frac{13}{2}x$, or $2y^2 + 2x^2 + 13x = 0$.

A first-order linear differential equation in the dependent variable $y$ and independent variable $x$ is an equation of the form: $a_1(x)\frac{dy}{dx} + a_0(x)y = b(x)$ (7.4) where $a_1(x)$ is not identically zero. Dividing equation (7.4) by $a_1(x)$ gives: $\frac{dy}{dx} + P(x)y = Q(x)$ (7.5) where $P(x) = \frac{a_0(x)}{a_1(x)}$ and $Q(x) = \frac{b(x)}{a_1(x)}$ are functions of $x$ or constants. Equation (7.5) is called the standard form of a first-order linear differential equation. Any first-order linear differential equation can be transformed into this standard form.

Example 1

Write each of the following first-order linear differential equations in standard form: (a) $x\frac{dy}{dx} = 3x - 2y$, $x > 0$ (b) $x^2y' = -y + \sin x$

Solution: (a) Given $x\frac{dy}{dx} = 3x - 2y$, $x > 0$. Divide both sides by $x$: $\frac{dy}{dx} = 3 - \frac{2y}{x}$ $\frac{dy}{dx} + \frac{2}{x}y = 3$ Comparing with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we have $P(x) = \frac{2}{x}$ and $Q(x) = 3$. Therefore, $\frac{dy}{dx} + \frac{2}{x}y = 3$ is the standard form. (b) Given $x^2y' = -y + \sin x$. Divide both sides by $x^2$: $y' = -\frac{y}{x^2} + \frac{\sin x}{x^2}$ $y' + \frac{1}{x^2}y = \frac{\sin x}{x^2}$ Comparing with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we have $P(x) = \frac{1}{x^2}$ and $Q(x) = \frac{\sin x}{x^2}$. Therefore, $y' + \frac{1}{x^2}y = \frac{\sin x}{x^2}$ is the standard form.

Solutions of the first-order linear differential equation: $\frac{dy}{dx} + P(x)y = Q(x)$ are obtained using the following steps:

1. Rearrange the given equation into the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x)$ and $Q(x)$ are functions of $x$ only.
2. Obtain the integrating factor $I(x)$, which has the form $I(x) = e^{\int P(x)dx}$.
3. Multiply both sides of the standard form equation by the integrating factor $I(x)$.
4. Integrate both sides with respect to $x$ and add a constant of integration $C$ to obtain the general solution.

Derivation of the Integrating Factor: Starting with $\frac{dy}{dx} + P(x)y = Q(x)$, multiplying by an integrating factor $R(x)$ gives: $R(x)\frac{dy}{dx} + R(x)P(x)y = R(x)Q(x)$ We want the left side to be the result of the product rule: $\frac{d}{dx}[R(x)y] = R(x)\frac{dy}{dx} + \frac{dR}{dx}y$ So, we require $R(x)P(x) = \frac{dR}{dx}$ $\frac{dR}{R} = P(x)dx$ $\int \frac{dR}{R} = \int P(x)dx$ $\ln|R| = \int P(x)dx$ $R(x) = e^{\int P(x)dx}$

Example 1

Find the general solution of each of the following differential equations: (a) $\frac{dy}{dx} + y = \sin x$ (b) $x\frac{dy}{dx} = 3x - 2y$, $x > 0$

Solution: (a) The equation is in standard form. $P(x) = 1$. Integrating factor: $I(x) = e^{\int 1 dx} = e^x$ Multiply the equation by $e^x$: $e^x\frac{dy}{dx} + e^xy = e^x\sin x$ $\frac{d}{dx}(e^xy) = e^x\sin x$ $e^xy = \int e^x\sin x dx$ Using integration by parts twice, we get: $\int e^x\sin x dx = \frac{1}{2}e^x(\sin x - \cos x) + C$ $e^xy = \frac{1}{2}e^x(\sin x - \cos x) + C$ $y = \frac{1}{2}(\sin x - \cos x) + Ce^{-x}$ (b) The standard form is $\frac{dy}{dx} + \frac{2}{x}y = 3$. $P(x) = \frac{2}{x}$. Integrating factor: $I(x) = e^{\int \frac{2}{x}dx} = e^{2\ln|x|} = e^{\ln(x^2)} = x^2$ Multiply the equation by $x^2$: $x^2\frac{dy}{dx} + 2xy = 3x^2$ $\frac{d}{dx}(x^2y) = 3x^2$ $x^2y = \int 3x^2 dx$ $x^2y = x^3 + C$ $y = x + \frac{C}{x^2}$

Example 2

Find the particular solution of each of the following initial-value problems: (a) $\frac{dy}{dx} + 4y = 2x$, $y(0) = 4$ (b) $(t + 1)s' + 2s = 3(t + 1) + (t + 1)^{-2}$, $s(0) = -2$

Solution: (a) $P(x) = 4$. Integrating factor: $I(x) = e^{\int 4 dx} = e^{4x}$ $e^{4x}\frac{dy}{dx} + 4e^{4x}y = 2xe^{4x}$ $\frac{d}{dx}(e^{4x}y) = 2xe^{4x}$ $e^{4x}y = \int 2xe^{4x}dx$ Using integration by parts: $\int 2xe^{4x}dx = \frac{1}{2}xe^{4x} - \frac{1}{8}e^{4x} + C$ $e^{4x}y = \frac{1}{2}xe^{4x} - \frac{1}{8}e^{4x} + C$ $y = \frac{1}{2}x - \frac{1}{8} + Ce^{-4x}$ Using $y(0) = 4$: $4 = -\frac{1}{8} + C \Rightarrow C = \frac{33}{8}$ $y = \frac{1}{2}x - \frac{1}{8} + \frac{33}{8}e^{-4x}$ (b) The standard form is $s' + \frac{2}{t+1}s = 3 + (t+1)^{-3}$. $P(t) = \frac{2}{t+1}$. Integrating factor: $I(t) = e^{\int \frac{2}{t+1}dt} = e^{2\ln|t+1|} = (t+1)^2$ $(t+1)^2s' + 2(t+1)s = 3(t+1)^2 + (t+1)^{-1}$ $\frac{d}{dt}((t+1)^2s) = 3(t+1)^2 + \frac{1}{t+1}$ $(t+1)^2s = \int (3(t+1)^2 + \frac{1}{t+1})dt$ $(t+1)^2s = (t+1)^3 + \ln|t+1| + C$ $s = (t+1) + \frac{\ln|t+1|}{(t+1)^2} + \frac{C}{(t+1)^2}$ Using $s(0) = -2$: $-2 = 1 + 0 + C \Rightarrow C = -3$ $s = (t+1) + \frac{\ln|t+1|}{(t+1)^2} - \frac{3}{(t+1)^2}$

Consider a differential equation of the form: $M(x, y)dx + N(x, y)dy = 0$ (7.8) If there exists a function $F(x, y) = k$, where $k$ is a constant, such that $dF = M(x, y)dx + N(x, y)dy = 0$, then the differential equation is exact, and $F(x, y) = k$ is a solution of equation (7.8). From the definition of the total differential: $dF = \frac{\partial F}{\partial x}dx + \frac{\partial F}{\partial y}dy = 0$ By comparison: $\frac{\partial F}{\partial x} = M(x, y)$ and $\frac{\partial F}{\partial y} = N(x, y)$ Assuming $F(x, y)$ has continuous second-order partial derivatives, then: $\frac{\partial}{\partial y}\left(\frac{\partial F}{\partial x}\right) = \frac{\partial^2 F}{\partial y \partial x} = \frac{\partial M}{\partial y}$ and $\frac{\partial}{\partial x}\left(\frac{\partial F}{\partial y}\right) = \frac{\partial^2 F}{\partial x \partial y} = \frac{\partial N}{\partial x}$ Since $F(x, y)$ has continuous second derivatives, $\frac{\partial^2 F}{\partial y \partial x} = \frac{\partial^2 F}{\partial x \partial y}$, which implies: $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ (7.9) Equation (7.9) is the condition for exactness of the differential equation $M(x, y)dx + N(x, y)dy = 0$.

Steps to solve exact differential equations:

1. Write the equation in the form $M(x, y)dx + N(x, y)dy = 0$.
2. Test for exactness using $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$.
3. If exact, integrate $M(x, y)$ with respect to $x$ or $N(x, y)$ with respect to $y$: $F(x, y) = \int M(x, y)dx + G(y)$ or $F(x, y) = \int N(x, y)dy + H(x)$ where $G(y)$ and $H(x)$ are functions of integration.
4. Differentiate $F(x, y)$ with respect to $y$ and equate it to $N(x, y)$ (or differentiate with respect to $x$ and equate to $M(x,y)$) to find $G'(y)$ (or $H'(x)$).
5. Integrate $G'(y)$ to find $G(y)$ (or integrate $H'(x)$ to find $H(x)$).
6. The solution is $F(x, y) = k$.

Example 1

Determine the exactness of each of the following differential equations: (a) $(x + 2y)dx + (y^2 - 2x)dy = 0$ (b) $(3x^2 + y\cos x)dx + (\sin x - 4y^3)dy = 0$ (c) $(2x^2 + y)dx + (x^2y - x)dy = 0$ (d) $(4y^5 + 3xy^2 + e^{xy})dx + (5x^4y + x^2y - 2x^2)dy = 0$

Solution: (a) $M(x, y) = x + 2y$, $\frac{\partial M}{\partial y} = 2$; $N(x, y) = y^2 - 2x$, $\frac{\partial N}{\partial x} = -2$. Since $\frac{\partial M}{\partial y} \ne \frac{\partial N}{\partial x}$, the equation is not exact. (b) $M(x, y) = 3x^2 + y\cos x$, $\frac{\partial M}{\partial y} = \cos x$; $N(x, y) = \sin x - 4y^3$, $\frac{\partial N}{\partial x} = \cos x$. Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, the equation is exact. (c) $M(x, y) = 2x^2 + y$, $\frac{\partial M}{\partial y} = 1$; $N(x, y) = x^2y - x$, $\frac{\partial N}{\partial x} = 2xy - 1$. Since $\frac{\partial M}{\partial y} \ne \frac{\partial N}{\partial x}$, the equation is not exact. (d) $M(x, y) = 4y^5 + 3xy^2 + e^{xy}$, $\frac{\partial M}{\partial y} = 20y^4 + 6xy + xe^{xy}$; $N(x, y) = 5x^4y + x^2y - 2x^2$, $\frac{\partial N}{\partial x} = 20x^3y + 2xy - 4x$. Since $\frac{\partial M}{\partial y} \ne \frac{\partial N}{\partial x}$, the equation is not exact.

Example 2

Solve each of the following differential equations: (a) $(2xy^3 + 2)dx + (3x^2y^2 + e^y)dy = 0$ (b) $x^2\frac{dy}{dx} = 1 - 2xy$

Solution: (a) $M(x, y) = 2xy^3 + 2$, $\frac{\partial M}{\partial y} = 6xy^2$; $N(x, y) = 3x^2y^2 + e^y$, $\frac{\partial N}{\partial x} = 6xy^2$. The equation is exact. $F(x, y) = \int (2xy^3 + 2)dx = x^2y^3 + 2x + G(y)$ $\frac{\partial F}{\partial y} = 3x^2y^2 + G'(y) = N(x, y) = 3x^2y^2 + e^y$ $G'(y) = e^y$ $G(y) = \int e^y dy = e^y$ Therefore, $F(x, y) = x^2y^3 + 2x + e^y = k$. (b) $(2xy - 1)dx + x^2dy = 0$. $M(x,y) = 2xy-1$, $\frac{\partial M}{\partial y} = 2x$. $N(x,y) = x^2$, $\frac{\partial N}{\partial x} = 2x$. The equation is exact. $F(x,y) = \int (2xy-1)dx = x^2y - x + G(y)$ $\frac{\partial F}{\partial y} = x^2 + G'(y) = N(x,y) = x^2$ $G'(y) = 0$ $G(y) = C$ (a constant) Therefore, $F(x,y) = x^2y - x = k$ or $x^2y - x = C$.

If a differential equation of the form $M(x, y)dx + N(x, y)dy = 0$ is not exact (i.e., $\frac{\partial M}{\partial y} \ne \frac{\partial N}{\partial x}$), it can sometimes be made exact by multiplying by an integrating factor $\mu(x, y)$. This leads to: $\frac{\partial (\mu M)}{\partial y} = \frac{\partial (\mu N)}{\partial x}$ (7.10) Which expands to: $\mu\frac{\partial M}{\partial y} + M\frac{\partial \mu}{\partial y} = \mu\frac{\partial N}{\partial x} + N\frac{\partial \mu}{\partial x}$ (7.11)

Case 1: If $\mu$ is a function of $x$ only, then $\frac{\partial \mu}{\partial y} = 0$, and equation (7.11) becomes: $\mu\frac{\partial M}{\partial y} = \mu\frac{\partial N}{\partial x} + N\frac{d\mu}{dx}$ $\frac{d\mu}{\mu} = \frac{1}{N}\left(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}\right)dx$ Integrating both sides: $\ln|\mu| = \int \frac{1}{N}\left(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}\right)dx$ $\mu(x) = e^{\int \frac{1}{N}\left(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}\right)dx}$ Let $f(x) = \frac{1}{N}\left(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}\right)$. Then, $\mu(x) = e^{\int f(x)dx}$.