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Chapter 10: Problem 9

Find a constant solution of \(y^{\prime}=t^{2} y-5 t^{2}\).

Short Answer

Expert verified
The constant solution is \( y = 5 \).

Step by step solution

01

Identify Constant Solution Conditions

A constant solution means that the derivative of the function is zero. Therefore, set the derivative to zero: \[ y' = 0 \]
02

Substitute into Given Equation

Substitute \( y' = 0 \) into the given differential equation: \[ 0 = t^2 y - 5 t^2 \]
03

Simplify the Equation

Divide both sides of the equation by \( t^2 \) (assuming \( t eq 0 \)): \[ 0 = y - 5 \]
04

Solve for y

Solve the simplified equation for \( y \): \[ y = 5 \]
05

Verify the Solution

Verify that substituting \( y = 5 \) into the original differential equation gives a true statement: \[ y' = t^2 y - 5 t^2 \] Substituting \( y = 5 \) yields: \[ y' = t^2 (5) - 5 t^2 \] \[ y' = 5 t^2 - 5 t^2 \] \[ y' = 0 \] This confirms that our solution \( y = 5 \) is correct.

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Key Concepts

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

constant solution
In solving the given differential equation, we start by finding a constant solution. A constant solution implies that the value of the function does not change over time. It remains static, unlike other dynamic solutions based on variable inputs. For a function to be constant, its derivative must be zero. This is because a non-changing value will have no rate of change over time. Hence, we set the derivative to zero ( y' = 0 ). By substituting this into the given differential equation, we can simplify and solve for y.
derivative
The concept of a derivative is crucial in understanding differential equations. A derivative represents the rate at which a function is changing at any given point. In mathematical notation, if we have a function y = f(t), the derivative, denoted as y' or dy/dt , shows how y changes with respect to t. For a constant solution, as discussed, this rate of change must be zero. In the given problem, starting with y' = 0 helps us focus on a scenario where y doesn't change, which simplifies the calculations significantly.
simplification
Simplification in mathematics involves making an equation or expression easier to work with. Once we set y' = 0 , we substitute it into the differential equation: 0 = t^2 y - 5 t^2 . To further simplify, we divide by t^2 (assuming t eq 0 ): 0 = y - 5 . Simplification helps us isolate variables and reach the solution more efficiently. In this case, it leads us directly to the value of y = 5 , which is our constant solution.
verification
Verification involves checking that our solution is indeed correct by substituting it back into the original equation. We found y = 5 as the constant solution. To verify, we substitute this back into the differential equation y' = t^2 y - 5 t^2 . Since y' = 0 for a constant solution, we get: 0 = t^2 (5) - 5 t^2 . Simplifying further: 0 = 5 t^2 - 5 t^2 confirms the validity of our solution. Hence, y = 5 satisfies the original equation.

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Most popular questions from this chapter

Solve the given equation using an integrating factor. Take \(t>0\). $$ y^{\prime}=e^{-t}(y+1) $$

One or more initial conditions are given for each differential equation in the following exercises. Use the qualitative theory of autonomous differential equations to sketch the graphs of the corresponding solutions. Include a \(y z\) -graph if one is not already provided. Always indicate the constant solutions on the \(t y\) -graph whether they are mentioned or not. $$ y^{\prime}=y-\frac{1}{4} y^{2}, y(0)=-1, y(0)=1 $$

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Suppose that \(f(t)\) is a solution of \(y^{\prime}=t^{2}-y^{2}\) and the graph of \(f(t)\) passes through the point \((2,3)\). Find the slope of the graph when \(t=2\).
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