Answer to Question #206788 in Differential Equations for Leela krishna

Question

X²(d²y/dx²)-4x(dy/dx)+6y=42/x⁴

Answer 1

First, we construct a general solution to the homogeneous equation:

${x^2}y'' - 4xy' + 6y = 0$

Let's make the substitution:

${x = {e^t},\;\;y' = {e^{ - t}}\frac{{dy}}{{dt}},}\;\; {y'' = {e^{ - 2t}}\left( {\frac{{{d^2}y}}{{d{t^2}}} - \frac{{dy}}{{dt}}} \right).}$

As a result, the homogeneous equation will take the form:

$\cancel{{{e^{2t}}}}\cancel{{{e^{ - 2t}}}}\left( {\frac{{{d^2}y}}{{d{t^2}}} - \frac{{dy}}{{dt}}} \right) - 4\cancel{{{e^t}}}\cancel{{{e^{ - t}}}}\frac{{dy}}{{dt}} + 6y = 0,\;\; \Rightarrow \frac{{{d^2}y}}{{d{t^2}}} - 5\frac{{dy}}{{dt}} + 6y = 0 = 0$

Let's solve the characteristic equation:

${k^2} - 5k + 6 = 0 \Rightarrow (k - 2)(k - 3) = 0 \Rightarrow {k_1} = 2,\,{k_2} = 3$

Then

${y_0}(t) = {C_1}{e^{2t}} + {C_2}{e^{3t}}$

We will seek a particular solution of the inhomogeneous equation $\frac{{{d^2}y}}{{d{t^2}}} - 5\frac{{dy}}{{dt}} + 6y = \frac{{42}}{{{e^{4t}}}} = 42{e^{ - 4t}}$ in the form

$Y = A{e^{ - 4t}} \Rightarrow Y' = - 4A{e^{ - 4t}} \Rightarrow Y'' = 16A{e^{ - 4t}}$

Then

$42A{e^{ - 4t}} = 42{e^{ - 4t}} \Rightarrow A = 1 \Rightarrow Y = {e^{ - 4t}}$

So,

$y(t) = {y_0}(t) + Y = {C_1}{e^{2t}} + {C_2}{e^{3t}} + {e^{ - 4t}} \Rightarrow y(x) = {C_1}{x^2} + {C_2}{x^3} + {x^{ - 4}}$

Answer: $y(x) = {C_1}{x^2} + {C_2}{x^3} + {x^{ - 4}}$