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Chapter 8: Problem 15

$$3 x y^{\prime \prime}+2(1-x) y^{\prime}-4 y=0$$

Short Answer

Expert verified
The given differential equation is a homogeneous second order ordinary differential equation (ODE). The solution involves classifying the ODE, solving for the homogeneous part using the auxiliary equation, and presenting the general solution which in this case equals the complementary function.

Step by step solution

01

Classify the Differential Equation

The given differential equation is \(3xy''+2(1-x)y'-4y = 0\). This equation can be classified as a variable coefficients linear second-order ordinary differential equation.
02

Identify Homogeneous and Non-homogeneous Parts

In this case, the differential equation is actually a homogeneous differential equation, meaning there is no function of just \(x\) on the right side of the equation.
03

Solve for Homogeneous Solution

The homogeneous linear equation is defined and solved by the auxiliary equation which is: \( ar^2+br+c=0 \), where \(a\), \(b\) and \(c\) are the coefficients of \(y'', y', y\) respectively. So plug in \(a=3x, b=2(1-x), c=-4\) into the auxiliary equation which gives us quadratic equation in terms of \(r\). Solving this equation will give roots, and those roots will give the complementary function using the formulas: If \(r1=r2\), so \(y_c(t)=c1*e^(r1*t) + c2*t*e^(r2*t)\); If \(r1鈮爎2\) both real, so \(y_c(t)=c1*e^(r1*t) + c2*e^(r2*t)\); If \(r1=a卤bi\), so \(y_c(t)=e^(a*t)[c1cos(b*t) + c2sin(b*t)]\).
04

Write down the general solution

The general solution of a linear differential equation is the sum of the complementary function and the particular integral. Since there is no non-homogeneous function in this problem, the general solution becomes the complementary function itself.

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Most popular questions from this chapter

In Problems \(29-34\), determine the Taylor series about the point \(x_{0}\) for the given functions and values of \(x_{0} .\) \(f(x)=\cos x, \quad x_{0}=\pi\)

\(9 t^{2} x^{\prime \prime}+9 t x^{\prime}+\left(9 t^{2}-4\right) x=0\)

$$\left(x^{2}-4\right) y^{\prime \prime}+(x+2) y^{\prime}+3 y=0$$

$$6 x^{3} y^{\prime \prime \prime}+\left(13 x^{2}-x^{3}\right) y^{\prime \prime}+x y^{\prime}-x y=0$$

$$6 x^{3} y^{\prime \prime \prime}+13 x^{2} y^{\prime \prime}+\left(x+x^{2}\right) y^{\prime}+x y=0$$
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