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$$x y^{\prime \prime}+y^{\prime}-4 y=0$$

Short Answer

Expert verified
The solution to the given differential equation is \(y = c_1x^2 + c_2x^{-2}\). The coefficients \(c_1\) and \(c_2\) will be determined by initial conditions, if provided.

Step by step solution

01

Form the Characteristic Equation

We have the differential equation \(xy'' + y' - 4y = 0\). The corresponding characteristic equation of this differential equation is given by \(mr^2 + r - 4=0 \) where \(m=1\)
02

Solve the Characteristic Equation

Solving the characteristic equation, we get \(r=(-1±\sqrt{1+16})/2\), which further simplifies to \(r_1=2\), \(r_2=-2\). The roots are real and distinct.
03

Obtain the General Solution

The general solution when the roots are real and distinct is given by \(y=c_1x^{r_1}+c_2x^{r_2}\), hence the solution will be \(y = c_1x^2 + c_2x^{-2}\).

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