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Chapter 4: Problem 7

Show that \(y=x-x^{-1}\) is a solution of the differential equation \(x y^{\prime}+y=2 x\)

Short Answer

Expert verified
Question: Show that the function \(y = x - x^{-1}\) is a solution to the differential equation \(x y' + y = 2x\). Answer: After calculating the derivative of the function with respect to x and substituting the function and its derivative into the given differential equation, we simplified the equation to show that both sides of the equation are equal. This confirms that the function \(y = x - x^{-1}\) is indeed a solution to the differential equation \(x y' + y = 2x\).

Step by step solution

01

Calculate the derivative of the function

To begin, we need to find the derivative of the given function \(y=x-x^{-1}\). We will do this using the power rule of differentiation. Rewrite the function as \(y = x^1 - x^{-1}\) to make it easier to calculate the derivative. \[ y' = \frac{d}{dx} (x^1 - x^{-1}) \] Using the power rule for differentiation, we have: \[ y' = 1 \cdot x^{1-1} - (-1) \cdot x^{-1-1} = x^0 + x^{-2} \] So, the derivative of the function is: \[ y'=1+x^{-2} \]
02

Substitute the function and its derivative into the differential equation

Now we will substitute the function \(y=x-x^{-1}\) and its derivative \(y'=1+x^{-2}\) into the given differential equation, \(x y' + y = 2x\). If the equation holds true, then our function is a solution to the differential equation. \[ x (1 + x^{-2}) + (x - x^{-1}) = 2x \]
03

Simplify the equation and check for validity

Now, simplify the equation to see if both sides are equal: \[ x + x^{-1} + x - x^{-1} = 2x \] As we can see, the \(x^{-1}\) terms cancel each other out, resulting in: \[ 2x = 2x \] The equation holds true, which means that \(y=x-x^{-1}\) is indeed a solution to the given differential equation \(x y' + y = 2x\).

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Key Concepts

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

Power Rule of Differentiation
Understanding the power rule of differentiation is crucial when dealing with polynomial functions. The rule states that to find the derivative of a term in the form of \( x^n \), you multiply by the exponent and then subtract one from the exponent. Formally, if \( f(x) = x^n \), then \( f'(x) = n \times x^{n-1} \).

For instance, when you have a function \( y = x^3 \), applying the power rule would give you a derivative of \( y' = 3x^{3-1} \), simplifying down to \( y' = 3x^2 \). It's a straightforward and efficient method that simplifies the process of differentiation, especially when dealing with polynomials.
Derivative Calculation
When calculating derivatives, it's essential to be methodical and apply the correct rules of differentiation. The derivative of a function at a point is the slope of the tangent line to the function at that point. It can also be thought of as the function's instantaneous rate of change.

To compute derivatives, you can use a variety of rules, including the power rule as discussed, the product rule, the quotient rule, and the chain rule, depending on the complexity of the function. For example, with a function like \( y = x^2 + 4x + 4 \), you would use the power rule to derive each term separately, resulting in \( y' = 2x + 4 \). If the function is more complex, involving multiplication or division of two functions, you would then apply the product or quotient rule, respectively.
Substitution Method in Differential Equations
Solving differential equations often requires clever techniques, one of which is the substitution method. This method involves replacing terms in a differential equation with a single variable to simplify and solve the equation more easily.

The substitution method is particularly useful when a differential equation includes a function and its derivative, as seen in the exercise provided. You would first calculate the derivative of the function (as we have done using the power rule), then replace both the function and its derivative into the differential equation. The goal of this replacement is to verify whether the function satisfies the equation or to solve for the function if it's not initially known.

Through substitution, we often reach a simplified equation that can be solved directly or further simplified to find the solution. It's a powerful tool in the mathematician's toolkit for making a complex differential equation more manageable.

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Most popular questions from this chapter

(a) Find the general solution \(y_{\mathrm{h}}\) of the homogeneous differential equation \(\frac{d y}{d x}+2 x y=0\) (b) Show that the general solution of the nonhomogeneous equation \(\frac{d y}{d x}+2 x y=3 e^{-x^{2}}\) is equal to the solution \(y_{b}\) in part (a) plus a particular solution to the nonhomogeneous equation.

Solve \(\left(x^{2}-y^{2}\right) d x+2 x y d y=0, y(1)=2\)

Find a curve each tangent of which forms with the coordinate axes a triangle of constant area \(\mathrm{S}=2 \mathrm{a}^{2}\).

Prove that a curve possessing the property that all its normals pass through a fixed point is a circle.

Solve the following differential equations: (i) \(y y^{\prime}+1=(x-1) e^{-y^{2} / 2}\) (ii) \(y^{\prime}+x \sin 2 y=2 x e^{-x^{2}} \cos ^{2} y\) (iii) \(y y^{\prime} \sin x=\cos x\left(\sin x-y^{2}\right)\) (iv) \(y^{\prime}=\frac{y^{2}-x}{2 y(x+1)}\)
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