Verifying Solutions To Differential Equations And Initial Value Problems

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Differential equations are the backbone of many scientific and engineering disciplines, describing the relationships between functions and their derivatives. When faced with a differential equation, a crucial task is to verify whether a given function is indeed a solution. Guys, let's dive into how we can rigorously verify solutions to these equations.

Verifying $y=2 e^{3 x}-2 x-2$ is a Solution to $y^{\prime}-3 y=6 x+4$

In this section, we aim to verify that the function $y=2 e^{3 x}-2 x-2$ is a solution to the differential equation $y^{\prime}-3 y=6 x+4$. To do this, we need to follow a systematic approach that involves finding the derivative of the given function, substituting both the function and its derivative into the differential equation, and then checking if the equation holds true. Let’s break it down step by step to make sure we understand each part clearly. Remember, the key is in the substitution and simplification!

First, we start with the given function:

y=2e3xβˆ’2xβˆ’2y=2 e^{3 x}-2 x-2

Next, we need to find the derivative of this function, denoted as $y^{\prime}$. Applying the rules of differentiation, we get:

yβ€²=ddx(2e3xβˆ’2xβˆ’2)y^{\prime} = \frac{d}{dx}(2 e^{3 x}-2 x-2)

Using the chain rule for the exponential term and the power rule for the linear term, we have:

yβ€²=2β‹…3e3xβˆ’2y^{\prime} = 2 \cdot 3 e^{3 x} - 2

Simplifying, we obtain:

yβ€²=6e3xβˆ’2y^{\prime} = 6 e^{3 x} - 2

Now that we have both $y$ and $y^{\prime}$, we can substitute them into the given differential equation:

yβ€²βˆ’3y=6x+4y^{\prime}-3 y=6 x+4

Substituting, we get:

(6e3xβˆ’2)βˆ’3(2e3xβˆ’2xβˆ’2)=6x+4(6 e^{3 x} - 2) - 3(2 e^{3 x}-2 x-2) = 6 x+4

Next, we need to simplify the left-hand side of the equation. Distribute the -3 across the terms inside the parenthesis:

6e3xβˆ’2βˆ’6e3x+6x+6=6x+46 e^{3 x} - 2 - 6 e^{3 x} + 6 x + 6 = 6 x + 4

Now, we combine like terms. Notice that the $6 e^{3 x}$ and $-6 e^{3 x}$ terms cancel each other out:

βˆ’2+6+6x=6x+4-2 + 6 + 6x = 6x + 4

This simplifies to:

6x+4=6x+46 x + 4 = 6 x + 4

Since the left-hand side of the equation equals the right-hand side, the equation holds true. Therefore, we have verified that the function $y=2 e^{3 x}-2 x-2$ is indeed a solution to the differential equation $y^{\prime}-3 y=6 x+4$. Awesome, right? We’ve just confirmed a solution!

In summary, the steps to verify a solution to a differential equation are:

  1. Find the derivative (or derivatives) of the given function.
  2. Substitute the function and its derivative(s) into the differential equation.
  3. Simplify the equation to see if it holds true.

If the equation holds true after simplification, then the given function is a solution. If not, then it is not a solution. Understanding and applying these steps is crucial in the study of differential equations, as it forms the basis for more complex problem-solving. So, keep practicing, and you'll become a pro at verifying solutions in no time!

Initial value problems (IVPs) add another layer of complexity to differential equations. An IVP consists of a differential equation along with an initial condition, which specifies the value of the function at a particular point. Verifying a solution to an IVP involves not only checking that the function satisfies the differential equation but also that it satisfies the initial condition. This process ensures that the function is a unique solution that fits the specific constraints of the problem. Let's see how we can verify solutions to IVPs with an example.

Verifying $y=2 e^{3 t}+4 ext{sin} t$ is a Solution to $y^{\prime}-2 y=4 ext{cos} t-8 ext{sin} t$ with Initial Condition

Let's tackle the task of verifying that $y=2 e^{3 t}+4 ext{sin} t$ is a solution to the initial-value problem (IVP)

y^{\prime}-3 y=4 ext{cos} t-8 ext{sin} t$ with $y(0) = 2$. This involves two key steps: first, we must confirm that $y=2 e^{3 t}+4 ext{sin} t$ satisfies the differential equation; second, we need to check if it meets the initial condition $y(0) = 2$. By completing both steps, we can confidently verify that the given function is a solution to the IVP. **Let's break it down and make sure we get this right!** ### Step 1: Verify the Differential Equation The first part of our task is to verify that $y=2 e^{3 t}+4 ext{sin} t$ satisfies the differential equation $y^{\prime}-3 y=4 ext{cos} t-8 ext{sin} t$. To do this, we need to find the derivative of the given function and then substitute both the function and its derivative into the differential equation. If the equation holds true, we've passed the first hurdle. We start with the given function: $y=2 e^{3 t}+4 ext{sin} t

Next, we find the derivative of this function, denoted as $y^{\prime}$. Applying the rules of differentiation, we get:

yβ€²=ddt(2e3t+4extsint)y^{\prime} = \frac{d}{dt}(2 e^{3 t}+4 ext{sin} t)

Using the chain rule for the exponential term and the basic derivative for the sine term, we have:

yβ€²=2β‹…3e3t+4extcosty^{\prime} = 2 \cdot 3 e^{3 t} + 4 ext{cos} t

Simplifying, we obtain:

yβ€²=6e3t+4extcosty^{\prime} = 6 e^{3 t} + 4 ext{cos} t

Now that we have both $y$ and $y^{\prime}$, we can substitute them into the differential equation:

yβ€²βˆ’2y=4extcostβˆ’8extsinty^{\prime}-2 y=4 ext{cos} t-8 ext{sin} t

Substituting, we get:

(6e3t+4extcost)βˆ’2(2e3t+4extsint)=4extcostβˆ’8extsint(6 e^{3 t} + 4 ext{cos} t) - 2(2 e^{3 t}+4 ext{sin} t) = 4 ext{cos} t-8 ext{sin} t

Next, we need to simplify the left-hand side of the equation. Distribute the -2 across the terms inside the parenthesis:

6e3t+4extcostβˆ’4e3tβˆ’8extsint=4extcostβˆ’8extsint6 e^{3 t} + 4 ext{cos} t - 4 e^{3 t} - 8 ext{sin} t = 4 ext{cos} t-8 ext{sin} t

Now, we combine like terms. The $e^{3t}$ terms combine to $2e^{3t}$.

2e3t+4extcostβˆ’8extsint2e^{3t} + 4 ext{cos} t - 8 ext{sin} t

Oops! It looks like there was a slight error in the original differential equation provided. The correct differential equation should be $y' - 3y$ instead of $y' - 2y$ to match the initial problem statement. Let's correct our substitution and simplification using the correct equation $y^{\prime}-3 y=4 ext{cos} t-8 ext{sin} t$

Substituting, we get:

(6e3t+4extcost)βˆ’3(2e3t+4extsint)=4extcostβˆ’8extsint(6 e^{3 t} + 4 ext{cos} t) - 3(2 e^{3 t}+4 ext{sin} t) = 4 ext{cos} t-8 ext{sin} t

Next, we need to simplify the left-hand side of the equation. Distribute the -3 across the terms inside the parenthesis:

6e3t+4extcostβˆ’6e3tβˆ’12extsint=4extcostβˆ’8extsint6 e^{3 t} + 4 ext{cos} t - 6 e^{3 t} - 12 ext{sin} t = 4 ext{cos} t-8 ext{sin} t

Now, we combine like terms. Notice that the $6 e^{3 t}$ and $-6 e^{3 t}$ terms cancel each other out:

4extcostβˆ’12extsint4 ext{cos} t - 12 ext{sin} t

Wait a minute! There seems to be another discrepancy. The correct equation after substitution and simplification should match the right-hand side, but we ended up with $4 ext{cos} t - 12 ext{sin} t$ instead of $4 ext{cos} t - 8 ext{sin} t$. This indicates that the given function $y=2 e^{3 t}+4 ext{sin} t$ does not satisfy the corrected differential equation $y^{\prime}-3 y=4 ext{cos} t-8 ext{sin} t$. It’s crucial to catch these errors and re-evaluate!

Step 2: Verify the Initial Condition

Even though we've found an issue with the differential equation part, let’s proceed with the second part to illustrate the process of verifying the initial condition. We need to check if the function $y=2 e^{3 t}+4 ext{sin} t$ satisfies the initial condition $y(0) = 2$. To do this, we substitute $t = 0$ into the function and see if the result equals 2.

Substituting $t = 0$ into the function, we get:

y(0)=2e3(0)+4extsin(0)y(0) = 2 e^{3 (0)}+4 ext{sin}(0)

Simplifying, we have:

y(0)=2e0+4(0)y(0) = 2 e^{0} + 4 (0)

Since $e^{0} = 1$ and $4(0) = 0$, the equation becomes:

y(0)=2(1)+0y(0) = 2 (1) + 0

y(0)=2y(0) = 2

The initial condition $y(0) = 2$ is satisfied by the function $y=2 e^{3 t}+4 ext{sin} t$. Great! We've confirmed the initial condition.

Conclusion

Although the function satisfies the initial condition, our analysis revealed that the given function $y=2 e^{3 t}+4 ext{sin} t$ does not satisfy the differential equation $y^{\prime}-3 y=4 ext{cos} t-8 ext{sin} t$. Therefore, the function is not a solution to the initial-value problem as originally stated. This highlights the importance of verifying both the differential equation and the initial condition to confirm a solution to an IVP.

In summary, to verify a solution to an initial value problem:

  1. Verify that the function satisfies the differential equation by substituting the function and its derivative(s) into the equation.
  2. Verify that the function satisfies the initial condition by substituting the given value of the independent variable into the function and checking if the result matches the given value of the dependent variable.

Both conditions must be met for the function to be a solution to the IVP. This comprehensive approach ensures that we have a unique solution that fits all the problem's requirements. Keep practicing these steps, and you'll be well-equipped to tackle any IVP verification!

Verifying solutions to differential equations and initial value problems is a fundamental skill in mathematics. By following a systematic approach, which includes finding derivatives, substituting into equations, and simplifying, we can confidently determine whether a given function is a solution. For initial value problems, it’s crucial to check both the differential equation and the initial condition to ensure a unique solution. With practice and attention to detail, you'll become proficient in verifying solutions and solving more complex problems in differential equations. Keep up the great work, and happy solving!