Math Problem Statement

میتونی توی حل این معادلات دیفرانسیل به کمک تبدیلات لاپلاس کمکم کنی

Solution

Sure, I can help you solve these differential equations using Laplace transforms. Let's break down each problem one by one.

Problem 4(a)

The differential equation is: y+3yy=xy'' + 3y' - y = x with initial conditions y(0)=1y(0) = 1 and y(0)=1y'(0) = -1.

  1. Apply the Laplace transform to both sides of the equation: L{y+3yy}=L{x}\mathcal{L}\{y'' + 3y' - y\} = \mathcal{L}\{x\}

  2. Use the properties of Laplace transforms: L{y}=s2Y(s)sy(0)y(0)\mathcal{L}\{y''\} = s^2Y(s) - sy(0) - y'(0) L{3y}=3(sY(s)y(0))\mathcal{L}\{3y'\} = 3(sY(s) - y(0)) L{y}=Y(s)\mathcal{L}\{y\} = Y(s)

  3. Substitute the initial conditions and the transforms into the equation: s2Y(s)s1(1)+3[sY(s)1]Y(s)=L{x}s^2Y(s) - s \cdot 1 - (-1) + 3[sY(s) - 1] - Y(s) = \mathcal{L}\{x\}

  4. Simplify the equation: (s2+3s1)Y(s)s+13=1s2(s^2 + 3s - 1)Y(s) - s + 1 - 3 = \frac{1}{s^2} (s2+3s1)Y(s)=1s2+s2(s^2 + 3s - 1)Y(s) = \frac{1}{s^2} + s - 2 Y(s)=1s21s2+3s1Y(s) = \frac{1}{s^2} \cdot \frac{1}{s^2 + 3s - 1}

  5. Solve for Y(s)Y(s): Decompose 1s2+3s1\frac{1}{s^2 + 3s - 1} into partial fractions and then use the inverse Laplace transform.

Problem 4(b)

The differential equation is: y+2y+5y=3exsin(x)y'' + 2y' + 5y = 3e^{-x} \sin(x) with initial conditions y(0)=0y(0) = 0 and y(0)=3y''(0) = 3.

  1. Apply the Laplace transform to both sides of the equation: L{y+2y+5y}=L{3exsin(x)}\mathcal{L}\{y'' + 2y' + 5y\} = \mathcal{L}\{3e^{-x} \sin(x)\}

  2. Use the properties of Laplace transforms: L{y}=s2Y(s)sy(0)y(0)\mathcal{L}\{y''\} = s^2Y(s) - sy(0) - y'(0) L{2y}=2(sY(s)y(0))\mathcal{L}\{2y'\} = 2(sY(s) - y(0)) L{5y}=5Y(s)\mathcal{L}\{5y\} = 5Y(s) L{3exsin(x)}=3(s+1)2+1\mathcal{L}\{3e^{-x} \sin(x)\} = \frac{3}{(s+1)^2 + 1}

  3. Substitute the initial conditions and the transforms into the equation: s2Y(s)sy(0)y(0)+2(sY(s)y(0))+5Y(s)=3(s+1)2+1s^2Y(s) - sy(0) - y'(0) + 2(sY(s) - y(0)) + 5Y(s) = \frac{3}{(s+1)^2 + 1}

  4. Simplify the equation: (s2+2s+5)Y(s)=3(s+1)2+1(s^2 + 2s + 5)Y(s) = \frac{3}{(s+1)^2 + 1} Y(s)=3((s+1)2+1)(s2+2s+5)Y(s) = \frac{3}{((s+1)^2 + 1)(s^2 + 2s + 5)}

  5. Solve for Y(s)Y(s): Decompose 3((s+1)2+1)(s2+2s+5)\frac{3}{((s+1)^2 + 1)(s^2 + 2s + 5)} into partial fractions and then use the inverse Laplace transform.

For detailed steps and solving, the partial fraction decomposition and inverse Laplace transform steps are needed.

Do you need detailed steps for the inverse Laplace transforms and partial fraction decomposition?

Suggested Questions

  1. How do you perform partial fraction decomposition?
  2. What are the properties of the Laplace transform?
  3. How do you find the inverse Laplace transform?
  4. How to solve a system of linear differential equations using Laplace transforms?
  5. How to handle non-homogeneous differential equations with Laplace transforms?

Tip

Remember that the initial conditions are crucial in using Laplace transforms for solving differential equations, as they simplify the transformed equation and make it easier to solve for Y(s)Y(s).

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Laplace Transforms
Partial Fraction Decomposition
Inverse Laplace Transform

Formulas

Laplace transform formula
Inverse Laplace transform formula
Partial fraction decomposition formula

Theorems

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Suitable Grade Level

Advanced undergraduate or graduate level