Question

Verify that the function is a solution of the differential equation on some interval, for any choice of the arbitrary constants appearing in the function. (a) \(y=c e^{2 x} ; \quad y^{\prime}=2 y\) (b) \(y=\frac{x^{2}}{3}+\frac{c}{x} ; \quad x y^{\prime}+y=x^{2}\) (c) \(y=\frac{1}{2}+c e^{-x^{2}} ; \quad y^{\prime}+2 x y=x\) (d) \(y=\left(1+c e^{-x^{2} / 2}\right) ;\left(1-c e^{-x^{2} / 2}\right)^{-1} \quad 2 y^{\prime}+x\left(y^{2}-1\right)=0\) (e) \(y=\tan \left(\frac{x^{3}}{3}+c\right) ; \quad y^{\prime}=x^{2}\left(1+y^{2}\right)\) (f) \(y=\left(c_{1}+c_{2} x\right) e^{x}+\sin x+x^{2} ; \quad y^{\prime \prime}-2 y^{\prime}+y=-2 \cos x+x^{2}-4 x+2\) (g) \(y=c_{1} e^{x}+c_{2} x+\frac{2}{x} ; \quad(1-x) y^{\prime \prime}+x y^{\prime}-y=4\left(1-x-x^{2}\right) x^{-3}\) (h) \(y=x^{-1 / 2}\left(c_{1} \sin x+c_{2} \cos x\right)+4 x+8 ;\) \(x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-\frac{1}{4}\right) y=4 x^{3}+8 x^{2}+3 x-2\)

Short answer

Question: Verify that the given functions are solutions of the specified differential equations on some interval for any choice of the arbitrary constants appearing in the functions. a) Function: y = ce^(2x), Differential Equation: y' = 2y Differentiate the function y with respect to x and verify the differential equation. b) Function: y = x^2/3 + c/x, Differential Equation: xy' + y = x^2 Differentiate the function y with respect to x and verify the differential equation.

Step-by-step solution

Differentiate y with respect to x

To find the first derivative, apply the chain rule: \(y' = \frac{d}{dx}(ce^{2x}) = 2ce^{2x}\). #Step 2: Verify the differential equation#

Substitute y' and y into the differential equation

Now, substitute the obtained \(y'\) and given y into the differential equation \(y' = 2y\): \(2ce^{2x} = 2(ce^{2x})\). Since the function satisfies the differential equation, it is a solution. #Part (b) y = x^2/3 + c/x, xy' + y = x^2# #Step 1: Differentiate the function#

Differentiate y with respect to x

Apply the chain rule and power rule to find the first derivative: \(y' = \frac{d}{dx}\left(\frac{x^2}{3} + \frac{c}{x}\right) = \frac{2x}{3} - \frac{c}{x^2}\). #Step 2: Verify the differential equation#

Substitute y' and y into the differential equation

Now, substitute the obtained \(y'\) and given y into the differential equation \(xy' + y = x^2\): \(x\left(\frac{2x}{3} - \frac{c}{x^2}\right) + \left(\frac{x^2}{3} + \frac{c}{x}\right) = x^2\). This simplifies to \(x^2 = x^2\), showing that the function satisfies the differential equation. The other parts will follow similar patterns of differentiation and verification. Make sure to carefully apply the relevant differentiation rules and simplify where necessary to confirm that the given functions and their derivatives satisfy the specified differential equations.

Key concepts

Verification of Solutions

In the context of differential equations, verifying a solution means checking whether a function satisfies the given differential equation. To verify a solution, follow these steps:

• Differentiate the given function according to the rules of calculus, such as the chain rule and power rule.
• Substitute the function and its derivative into the differential equation.
• Simplify the equation to check if both sides are equal. If they are, the function is indeed a solution.

This process helps ensure that the function meets all conditions of the differential equation for the interval and arbitrary constants provided. Understanding this verification ensures that solutions are not blindly accepted but scientifically validated.

Arbitrary Constants

Arbitrary constants are terms added to the general solution of a differential equation. They account for the infinite number of particular solutions that satisfy the equation. These constants can take any value, depending on the initial or boundary conditions.

• They appear as parameters (like the constant 'c' in a solution).
• In practice, they are determined by additional conditions, such as specific values of the solution at given points.

Recognizing arbitrary constants in a solution is crucial as it signifies the flexibility and adaptability of a differential equation to various scenarios. Identifying and calculating these constants play a vital role in tailoring the solution to specific problems.

Chain Rule

The chain rule is a fundamental differentiation technique used when dealing with composite functions. A composite function is a function composed of other functions. The chain rule provides a way to differentiate these functions efficiently:

• If you have a function of the form \(f(g(x))\), where \(f\) and \(g\) are functions, the derivative is given by \(f'(g(x)) \cdot g'(x)\).
• This method breaks down the differentiation process into manageable parts, making it easier to apply to more complex functions.

In differential equations, the chain rule is often used to differentiate functions implicitly dependent on other variables. Grasping this rule helps in tackling a broad range of dynamic problems regarding rates of change.

Power Rule

The power rule is one of the simplest and most common rules for differentiation. It is applicable when dealing with functions that involve powers of a variable. Here’s how it works:

• If you have a function \(y = x^n\), the derivative is \(y' = nx^{n-1}\).
• This rule provides a quick method to find the slope of a line tangent to the curve at any given point.

The power rule is frequently utilized in solving differential equations due to its simplicity and effectiveness. By using the power rule, complex polynomial expressions can be differentiated without difficulty, making it a critical tool in calculus.