Solve the Given Differential Equation by Variation of Parameters. xy'' − 4y' = x⁴ 

We will be solving this by finding the homogeneous and complementary solutions of the equation.

Answer: The general solution of differential equation xy'' − 4y' = x⁴ by variation of parameters is y = \(c_1\)+ \(c_2\)x⁵  - 1/25 x⁵ + 1/5 x⁵ lnx.

Let's solve this step by step.

Explanation:

Given, differential equation: xy'' − 4y' = x⁴ 

Multiply by x on both sides of xy'' − 4y' = x⁴ :

x²y'' − 4xy' = x⁵ 

Let's solve the corresponding homogeneous differential equation: x²y'' − 4xy' = 0

Let, y = x^m 

⇒ y' = mx^m-1 --------------- (1)

⇒ y'' = m(m - 1)x^m-2 --------------- (2)

Substitue (1) and (2) in x²y'' − 4xy' = 0,

⇒ x²m(m - 1)x^m-2 − 4xmx^m-1 = 0

⇒ m(m - 1)x^m − 4mx^m = 0

⇒ x^m [m(m - 1) − 4m] = 0

⇒ [m(m - 1) − 4m] = 0

⇒ m² - m − 4m  = 0

⇒ m² − 5m  = 0

⇒ m(m - 5) = 0

⇒ m = 0, 5

The two solutions are y\(_1\) = x^m = x⁰ = 1 and y\(_2\) = x^m = x⁵ 

Therefore, the complimentary solutions y\(_c\) is y\(_c\) = \(c_1y_1 + c_2y_2\)

y\(_c\) = \(c_1\)+ \(c_2\)x⁵ 

Now, find the remaining particular solution y\(_p\).

x²y'' − 4xy' = x⁵ 

Dividing by x² on both the sides of x²y'' − 4xy' = x⁵, we get:

y'' − (4/x) y' = x³ 

The equation is in the form of y'' + P(x)y' + Q(x) = g(x),

On comparing we get g(x) = x³ 

Wronskian formula: W(f, g) = f g' - g f'

W(y\(_1\), y\(_2\)) = y\(_1\) y\(_2\) ' - y\(_2\) y\(_1\) ' = 1(5x⁴) - x⁵(0) = 5x⁴ = W(x)

\(y_p = -y_1\int \dfrac{y_2(x)g(x)}{W(x)}dx +y_2\int \dfrac{y_1(x)g(x)}{W(x)} dx\)

\(y_p = -1\int \dfrac{x^5.x^3}{5x^4}dx +x^5\int \dfrac{1.x^3}{5x^4}dx \)

⇒ \(y_p\) = (-1/25 x⁵)(1) + (1/5 lnx)(x⁵)

⇒ \(y_p\) = -1/25 x⁵ + (1/5) x⁵ lnx

⇒ y = \(c_1\)+ \(c_2\)x⁵  - 1/25 x⁵ + (1/5) x⁵ lnx

Thus, the general solution of xy'' − 4y' = x⁴  is y = \(c_1\)+ \(c_2\)x⁵  - 1/25 x⁵ + 1/5 x⁵ lnx.