Math Problem Statement

Show that the equations y(x) are solution to the Ordinary Differential Equations.

  1. ax + y=0 y(x) = Ce=*
  2. y' = xy y(x) =Cel

Solution

Let's analyze both cases and verify whether the given functions y(x)y(x) are indeed solutions to the corresponding ordinary differential equations (ODEs).

Problem 2:

The given ODE is: ax+y=0a x + y = 0 The proposed solution is: y(x)=Cexy(x) = C e^{x}

Step-by-Step Verification:

  1. First, substitute y(x)=Cexy(x) = C e^{x} into the ODE.

The ODE becomes: ax+Cex=0a x + C e^{x} = 0

  1. Since this equation needs to hold for all xx, we see that it cannot be true for arbitrary xx and constant CC, because the terms axa x and CexC e^x are not comparable or cancelable. Thus, the proposed solution y(x)=Cexy(x) = C e^x is not a solution to this ODE.

Conclusion:

The function y(x)=Cexy(x) = C e^x does not satisfy the equation ax+y=0ax + y = 0.


Problem 3:

The given ODE is: y=xyy' = xy The proposed solution is: y(x)=Cex22y(x) = C e^{\frac{x^2}{2}}

Step-by-Step Verification:

  1. Differentiate y(x)=Cex22y(x) = C e^{\frac{x^2}{2}} with respect to xx.

y(x)=ddx(Cex22)=Cex22ddx(x22)=Cex22xy'(x) = \frac{d}{dx} \left( C e^{\frac{x^2}{2}} \right) = C \cdot e^{\frac{x^2}{2}} \cdot \frac{d}{dx} \left( \frac{x^2}{2} \right) = C \cdot e^{\frac{x^2}{2}} \cdot x

Thus, we have: y(x)=xCex22y'(x) = x C e^{\frac{x^2}{2}}

  1. Now, substitute y(x)=Cex22y(x) = C e^{\frac{x^2}{2}} and y(x)=xCex22y'(x) = x C e^{\frac{x^2}{2}} into the original ODE y=xyy' = xy.

xCex22=xCex22x C e^{\frac{x^2}{2}} = x \cdot C e^{\frac{x^2}{2}}

Both sides of the equation are equal, which confirms that the proposed solution is indeed correct.

Conclusion:

The function y(x)=Cex22y(x) = C e^{\frac{x^2}{2}} is a solution to the ODE y=xyy' = xy.


Summary:

  1. For ax+y=0ax + y = 0, the function y(x)=Cexy(x) = C e^x is not a solution.
  2. For y=xyy' = xy, the function y(x)=Cex22y(x) = C e^{\frac{x^2}{2}} is a solution.

Would you like further details on this, or do you have any questions?

Here are 5 related questions to explore:

  1. How do you verify if a function is a solution to an ODE in general?
  2. What are the general methods for solving first-order linear ODEs?
  3. How does separation of variables work for solving ODEs?
  4. Can we solve second-order differential equations with a similar method?
  5. How does the constant CC arise in the solutions of these ODEs?

Tip: When solving ODEs, always differentiate the proposed solution and substitute back into the equation to verify correctness.

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Math Problem Analysis

Mathematical Concepts

Ordinary Differential Equations (ODEs)
First-order Linear ODEs
Exponential Functions
Differentiation

Formulas

ODE: ax + y = 0
Solution Attempt: y(x) = C e^x
ODE: y' = xy
Solution: y(x) = C e^(x^2/2)

Theorems

Verifying solutions to ODEs by substitution

Suitable Grade Level

Undergraduate - First-year Calculus or Differential Equations