How do you find the general solution of a differential equation

Steps for Finding General Solutions to Differential Equations Using Antidifferentiation

Step 1: Rewrite the given differential equation in the form {eq}f(y)\text{d}y = g(x)\text{d}x {/eq}; that is, rewrite the given equation so that all occurrences of {eq}y {/eq} are on one side of the equation and the {eq}x {/eq}'s on the other side of the equation.

Step 2: If necessary, simplify your work after Step 1 before integrating.

Step 3: Integrate both sides of the equation. That is, evaluate both of the integrals in the following equation: {eq}\int f(y)\text{d}y = \int g(x)\text{d}x + C {/eq}.

Step 4: Simplify both the left and right-hand side of the equation after Step 3, and solve the equation for the function {eq}y {/eq} as a function of {eq}x {/eq}. That is, express the final answer in the form {eq}y=f(x) + C {/eq}.

Step 5: (This step only applies if, in addition to a given differential equation, we are given an initial condition of the form {eq}f(x_0)=y_0 {/eq}). If an initial condition {eq}f(x_0)=y_0 {/eq} is given, then use this information to find the value {eq}C {/eq} from Step 4, and then update your expression in Step 4.

Vocabulary for Finding General Solutions to Differential Equations Using Antidifferentiation

• Separation of Variables: This is the method of solving a differential equation provided by the step-by-step procedure above.
• Antidifferentiation: By the Fundamental Theorem of Calculus, the antiderivative (as a function) is the inverse function of the derivative. If {eq}y=f(x) {/eq}, then it is common to write the antiderivative as

{eq}y=F(x)+C {/eq}.

• Initial condition: Given a differential equation, and initial condition is a side condition of the form {eq}y(x_0) = y_0 {/eq}.

Now that we have reviewed a step-by-step procedure for solving differential equations by using separation of variables and antidifferentiation, let's hone our skills by working through three examples.

Example 1 - Finding General Solutions to Differential Equations Using Antidifferentiation

Use antidifferentiation to determine the general solution to the differential equation {eq}\dfrac{dy}{dx}=\dfrac{6x}{y+2} {/eq}.

Step 1: Rewrite the given differential equation in the form {eq}f(y)\text{d}y = g(x)\text{d}x {/eq}.

We can write the given differential equation in the form {eq}(y+2)\text{d}y = 6x \text{d} x {/eq}.

Step 2: If necessary, simplify your work after Step 1 before integrating.

The expressions in Step 1 are already simplified, so let's proceed to the next step.

Step 3: Integrate both sides of the equation.

We have {eq}\int (y+2)\text{d}y =\int 6x \text{d}x {/eq}. Both of these integrals can be solved using the power rule; this leads to {eq}y^{2}/2 + 2y = 6x{2}/2+C. {/eq}, which simplifies to {eq}y^{2}/2 + 2y = 3x{2}+C. {/eq}.

Step 4: Solve for {eq}y {/eq} as a function of {eq}x {/eq}.

From Step 3, we have {eq}y^{2}+2y = 3x^{2} + C {/eq}, and this can be viewed as a quadratic in the function {eq}y {/eq}; that is, we can use the quadratic formula on the equation {eq}y^{2}/2 + 2y - (3x^{2}+C) = 0 {/eq} in order to obtain {eq}y=-2\pm \frac{2^2 - 4\cdot \frac{1}{2}\cdot (-(3x^{2}+C))}{2\cdot \frac{1}{2}} {/eq}. This can be rewritten as {eq}y=-2 \pm 2\sqrt{1+\frac{3x^{2}+C}{2}} {/eq}.

Example 2 - Finding General Solutions to Differential Equations Using Antidifferentiation

Find the solution to the differential equation {eq}\frac{\text{d}y}{\text{d}x} = 4x^{3}e^{-y} {/eq} subject to the initial condition {eq}y(1)=3 {/eq}.

Step 1: Rewrite the given differential equation in the form {eq}f(y)\text{d}y = g(x)\text{d}x {/eq}.

Multiply both sides of the equation by {eq}\text{dx} {/eq} and divide both sides of the equation by {eq}e^{-y} {/eq} in order to obtain {eq}e^{y}\text{d}y =4x^{3}\text{d}x {/eq}.

Step 2: If necessary, simplify your work after Step 1 before integrating.

There is no simplifying necessary in this case, so let's proceed to integrate in Step 3.

Step 3: Integrate both sides of the equation.

The integral on the left-hand side of the equation in Step 1 is {eq}\int e^{y}\text{d}y = -e^{-y} {/eq} (once again, we'll add the constant {eq}C {/eq} to the other side); by using the power rule, the integral of the right-hand side of the equation in Step 1 is {eq}\int 4x^{3}\text{d}x=x^{4}+C {/eq}.

Step 4: Solve for {eq}y {/eq}.

To solve the equation {eq}e^{y} = x^{4}+C {/eq} for {eq}y {/eq}, let's take a natural logarithm of both sides to obtain {eq}y = \ln (x^4 + C) {/eq}.

Step 5: Use the initial condition to solve for {eq}C {/eq} and then update your expression for the function {eq}y {/eq}.

We have {eq}y(1) = 3 {/eq}, so plugging this into our equation obtained in Step 4, we have {eq}3 = \ln (1^4+C) = \ln(1+C) {/eq}, so solve for {eq}C {/eq} to obtain {eq}C=e^{3}-1 {/eq}. Plugging this value of {eq}C {/eq} back into our expression obtained in Step 4 gives us {eq}y=\ln (x^{4}+e^{3}-1) {/eq}.

Example 3 - Finding General Solutions to Differential Equations Using Antidifferentiation

Use antidifferentiation to determine the general solution to the differential equation {eq}\dfrac{dy}{dx}=\dfrac{2^y}{x^2+1} {/eq}.

Step 1: Rewrite the given differential equation in the form {eq}f(y)\text{d}y = g(x)\text{d}x {/eq}.

Divide both sides of the given differential equation by {eq}\text{d}x {/eq} and divide both sides of the equation by {eq}2^{y} {/eq} in order to obtain {eq}2^{-y}\text{d}y=\frac{\text{d}x}{x^{2}+1} {/eq}.

Step 2: If necessary, simplify your work after Step 1 before integrating.

There is no need to simplify, so let's proceed by integrating in Step 3.

Step 3: Integrate both sides of the equation.

The integral of the left-hand side is {eq}\int 2^{-y}\text{d}y = -\frac{2^{-y}}{\ln 2} {/eq} (it may help to use a U-substitution {eq}u = -y {/eq} when evaluating this integral). The integral of the right-hand side in Step 1 is {eq}\int \frac{\text{d}x}{x^{2}+1} = \text{arctan} (x) + C {/eq}.

Step 4: Solve for {eq}y {/eq}.

To solve for {eq}y {/eq}, let's multiply both sides of the equation {eq}-\frac{2^{-y}}{\ln 2} = \text{arctan}(x) + C {/eq} by -1, and then multiply both sides by {eq}\ln 2 {/eq} to obtain {eq}2^{-y} = - \ln 2 (\arctan (x) + C) {/eq}. Take a base 2 logarithm of both sides, and then multiply both sides by -1 to obtain {eq}y= -\log _{2} (-\ln 2 (\arctan (x) + C) {/eq}.

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