Question

1. y'' - 2y' = 0                ans, y=c₁+c₂e^2x
2. 4y'' - 4y' + y = 0       ans, y=c₁e^1/2x+c₂xe^1/2x

Answer 1

y'' - 2y' = 0  

This is linear homogeneous of second order with with constant coefficients. For solution we form characteristic algebraic equation

$k^2-2\cdot k=0$

Or $k\cdot(k-2)=0$

From it we have two roots:

$k_1=0,k_2=2$

These roots are real and different

Therefore fundamental set of solutions of DE is

$\{e^{k_1\cdot x}, e^{k_2\cdot x}\}=$

$\{e^{k_1\cdot x}, e^{k_2\cdot x}\}=\{e^{0\cdot x}, e^{2\cdot x} \}=\{1, e^{2\cdot x} \}$

General solution of DE is a linear combination of fyndamental solutions, thus $y(c_1,c_2,x)=c_1\cdot y_1(x)+c_2\cdot y_2(x)=c_1\cdot 1+c_2\cdot e^{2\cdot x}=c_1+c_2\cdot e^{2\cdot x},c_1,c_2\in R -any\space numbers$

4y'' - 4y' + y = 0

This is linear homogeneous of second order with with constant coefficients. For solution we form characteristic algebraic equation

$4\cdot k^2-4\cdot k+1=0$

This is the quadratic equation

It's roots are:

$k_{1,2}=\frac{4\pm\sqrt{4^2-4\cdot4\cdot1}}{2\cdot 4}=\frac{4\pm 0}{2\cdot 4}=\frac{1}{2}$

Thus we have only one root k=2 but with multiplicity equals to 2.

Then fundamental set of solutions of the DE is$\{e^{k\cdot x},x\cdot e^{k\cdot x}\}=\{e^{\frac{1}{2}\cdot x},x\cdot e^{\frac{1}{2}\cdot x}\}$ .

General solution of DE is a linear combination of fyndamental solutions, thus

$y(c_1,c_2,x)=c_1\cdot y_1(x)+c_2\cdot y_2(x)=c_1\cdot e^{\frac{1}{2}\cdot x} +c_2\cdot x\cdot e^{\frac{1}{2}\cdot x},=e^{\frac{1}{2}\cdot x}\cdot (c_1+x\cdot c_2),\space c_1,c_2\in R -any\space numbers$