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Solve the initial-value problem. $$ y^{\prime}=2(10-y), y(0)=1 $$

Short Answer

Expert verified
The solution to the initial value problem is \[ y = 20 - 19 e^{-x} \]

Step by step solution

01

Identify the type of differential equation

This is a first-order linear differential equation.
02

Rewrite the equation in standard form

Rewrite the given differential equation in the standard form: \[y' + ay = b\] In our case, it turns into: \[y' + y = 20\]
03

Find the integrating factor

The integrating factor for a first-order linear differential equation is:\[ \text{IF} = e^{\frac{\text{integral of } a \text{ with respect to } x}} \] Here, a=1, so: \[ \text{IF} = e^{\text{integral of } 1 \text{ with respect to } x} = e^x \]
04

Multiply through by the integrating factor

Multiply the entire differential equation by the integrating factor \(e^x\):\[ e^x y' + e^x y = 20 e^x\]
05

Simplify the equation

Notice that the left side is now the derivative of \(e^x y\): \[ \frac{d}{dx}(e^x y )= 20 e^x\]
06

Integrate both sides

Integrate both sides with respect to x: \[ \text{integral of } \frac{d}{dx}(e^x y ) = \text{integral of } 20 e^x dx \] \[ \text{which simplifies to: } e^x y = 20 e^x +C \]
07

Solve for y

Divide both sides by \(e^x\): \[ y = 20 + Ce^{-x}\]
08

Apply initial condition

Use the initial condition \( y(0)=1\) to find C: \[ 1 = 20 + C e^0 \] Simplifying it: \[ 1 = 20 + C \]Thus, \[ C = 1-20 \] \[ C = -19 \]
09

Write the final solution

Substitute C back into the solution: \[ y = 20 - 19 e^{-x} \] Final Solution: \[ y = 20 - 19 e^{-x} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

differential equations
A differential equation is a mathematical equation that relates some function with its derivatives. In simple terms, it tells us how a function changes over time. Differential equations are an essential tool in modeling real-world systems in science and engineering. They come in various shapes and complexities but are mainly categorized into ordinary differential equations (ODEs) and partial differential equations (PDEs). Our problem involves an ordinary differential equation since it includes only ordinary derivatives of the function.
first-order linear differential equations
A first-order linear differential equation is a type of differential equation that can be written in the form \[y' + a(x)y = b(x)\] where
  • \(y'\) is the derivative of \(y\) with respect to \(x\)
  • \(a(x)\) and \(b(x)\) are functions of \(x\).
In our given exercise, the equation \[y' = 2(10 - y)\] can be rewritten as \[y' + y = 20\] which fits the standard form. Here, \(a(x) = 1\) and \(b(x) = 20\). This kind of equation is straightforward to solve using an integrating factor, which leads us to our next core concept.
integrating factor
The integrating factor is a handy tool for solving first-order linear differential equations. The integrating factor, usually denoted as \(IF\), is a function used to simplify the equation and make it easier to solve. It is given by: \[ IF = e^{\text{integral of } a(x) \text{ with respect to } x} \] For our problem, where \(a(x) = 1\), the integrating factor becomes: \[IF = e^{\text{integral of } 1 \text{ with respect to } x} = e^x \] Multiplying the entire differential equation by \(e^x\) simplifies the left-hand side into the derivative of \(e^x y\), allowing us to integrate both sides easily. This step eventually helps us solve for \(y\) and find the particular solution by applying the initial condition \(y(0) = 1\). Integrating factors make complex differential equations much more manageable.

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Most popular questions from this chapter

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