Question

find the general solution of the differential equation x²y"-4xy'+6y=21x-⁴

Answer 1

x
${x}^{2} y - 4xy+ 6y = 21 {x4}^{4}$

Answer 2

The general solution of the differential equation is $y = Ax^{3} + B x^{2} + \frac{1}{2x^{4}}$

Step-by-step explanation:

Given: Differential equation $x^{2} y'' -4xy' + 6y = 21x^{-4}$

To Find: Solution of the given differential equation

Solution:

• Finding the general solution of the given differential equation

Considering the differential equation $x^{2} y'' -4xy' + 6y = 21x^{-4}$ such that put $x = e^z$, and, $xy' = Dy$ & $x^{2} y'' = D (D-1)y$ where $D = d/dz$.

Therefore, the differential equation becomes,

$\Rightarrow (D (D-1) - 4D + 6)y = 21e^{-4z}$

$\Rightarrow (D^2 - 5D + 6)y = 21e^{-4z}$

To find the complementary factor, we have the above differential equation such that:

$\Rightarrow m^2 -5m + 6 = 0 \Rightarrow m = 3, 2$

$\therefore y_{_{CF}}=A e^{3z} + B e^{2z}$

And, the particular integral is determined by;

$y_{_{PI}}= \dfrac{1}{D^2 - 5D + 6} (21e^{-4z})$

Using the rule $\frac{1}{f(D)}e^{ax} = \frac{1}{f(a)}e^{ax}$ in the above expression, we get;

$y_{_{PI}}=21 \dfrac{1}{(-4)^2 - 5(-4) + 6} e^{-4z} = \dfrac{e^{-4z} }{2}$

The general solution of the differential equation is $y = y_{_{CF}} + y_{_{PI}} = A e^{3z} + B e^{2z} + \dfrac{e^{-4z} }{2}$

Since $x = e^z$, then,$y = A x^{3} + B x^{2} + \frac{1}{2x^{4}}}$

Hence, The general solution of the differential equation is $y = Ax^{3} + B x^{2} + \dfrac{1}{2x^{4}}$