Question

find a power series solution of xy'=y​

Answer 1

Answer:

You can generate the Taylor series easily enough. Assume y(0)=a and y’(0)=b, and y(x)=sum(c[n]*x^n,n,0,inf). We can compute the derivatives at x=0 directly from the ODE and its derivatives, so y’’(0)=0*y(0)=0*a=0, then we get y’’’=x*y’+y so y’’’(0)=0*y’(0)+y(0)=a, then y’’’’=x*y’’+y’+y’=x*xy+2*y’=x^2*y+y’ so y’’’’(0)=0^2*y(0)+y’(0)=b. and so on, giving y=a+b*x+a*x^3/6+b*x^4/24+… using diff(y(x),x,m) at x=0 is m!*c[m]

Answer 2

Final answer: y = a₁ x

         y = a₁ x, where a₁ ∈ R

Given: We are given x y' = y​

To find: We have to find a power series solution of x y' = y​

Explanation:

• Let y = ∑ ₙ₋₀ to ∞ (aₙ xⁿ)
• Then, y′ = ∑ ₙ₋₁ to ∞ (n aₙ xⁿ⁻¹)
• Therefore, x y′ = y ⇒ x ∑ ₙ₋₁ to ∞ (n aₙ xⁿ⁻¹) = ∑ ₙ₋₀ to ∞ (aₙ xⁿ)
• ⇒ ∑ ₙ₋₁ to ∞ (n aₙ xⁿ) = ∑ ₙ₋₀ to ∞ (aₙ xⁿ)
• ⇒ ∑ ₙ₋₀ to ∞ (aₙ xⁿ) - ∑ ₙ₋₁ to ∞ (n aₙ xⁿ) = 0
• ⇒ a₀ + ∑ ₙ₋₁ to ∞ (aₙ xⁿ) - ∑ ₙ₋₁ to ∞ (n aₙ xⁿ) = 0
• ⇒ a₀ + ∑ ₙ₋₁ to ∞ (aₙ (1-n) xⁿ) = 0
• ⇒ ∑ ₙ₋₀ to ∞ (aₙ (1-n) xⁿ) = 0
• ⇒ ∑ ₙ₋₀ to ∞ (aₙ (1-n) xⁿ) = ∑ ₙ₋₀ to ∞ (0 × xⁿ)
• ⇒ aₙ (1-n) = 0, for n ∈ N
• ⇒ aₙ = (0/(1-n)) = 0, if n ≠ 1
• ⇒ a₁ ∈ R
• Hence, y = ∑ ₙ₋₀ to ∞ (aₙ xⁿ) = a₀ + a₁ x + a₂ x² + ...... = a₁ x
• Therefore, the series solution to the given equation is :

              y = a₁ x, where a₁ ∈ R.

To know more about the concept please go through the links;

https://brainly.in/question/2229138

https://brainly.in/question/13614193

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