Question

x^2d^2y/dx^2+2xdy/dx-20y=(x+1)^2​

Answer 1

The general solution for the differential equation is:

y(x) = C₁ × x⁻⁵+ C₂ × x⁴

Given:

x²d²y/dx²+ 2xdy/dx- 20y = (x+1)²

To find:

Solve the given equation

Solution:

The given equation is a second-order linear homogeneous differential equation with variable coefficients.

Let's solve it step by step.

The equation is:

x²d²y/dx² +2xdy/dx -20y = (x+1)²

Rewrite the equation in standard form

To simplify the equation, let's divide through by x²:

(d²y/dx²) + (2/x) × (dy/dx) - (20/x²) × y = (x + 1)² / x²

Assume a solution of the form y = $x^{r}$

Differentiate y with respect to x:

dy/dx = r × $x^{r-1}$

Differentiate again:

(d²y/dx²) = r × (r - 1) × $x^{r-2}$

Substitute the assumed solution into the differential equation

Substituting the derivatives and y into the equation, we get:

r × (r - 1) × $x^{r-2}$ + (2/x) × r × $x^{r-1}$ - (20/x²) * $x^{r}$ = (x + 1)² / x²

Simplifying further:

r × (r - 1) × $x^{r-2}$ + 2r × $x^{r-2}$ - 20 × $x^{r-2}$ = (x + 1)² / x²

Solve for the value of r

To solve for r, we equate the coefficients of like powers of x to zero:

r * (r - 1) + 2r - 20 = 0

Simplifying the equation:

r² - r + 2r - 20 = 0

r² + r - 20 = 0

Factoring the quadratic equation:

(r + 5)(r - 4) = 0

So, we have two possible values for r:

r₁ = -5 and r₂ = 4

Find the general solution

Since we have two distinct roots, the general solution for the differential equation is:

y(x) = C₁ × x⁻⁵+ C₂ × x⁴

Where C₁ and C₂ are constants to be determined based on initial conditions or additional information.

Therefore,

The general solution for the differential equation is:

y(x) = C₁ × x⁻⁵+ C₂ × x⁴

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