Question

Question 4 < > Solve the initial value problem below. x+y'' - xy' + y = 0, y(1) = – 5, y'(1) = 0 y

Answer 1

given initial value problem x²y’’ - xy’ + y = 0 , y(1) = -5 and Y’(1) = 0

so this was in the form of   ax²y’’ + bxy’ + cy = 0 which was known as cauchy – euler homogeneous differential equation.

So the general solution is in the form of Y = Yc + Yp

Yc = homogeneous solution

Yp = particular solution

So for Yc we need to take homogeneous part   x²y’’ - xy’ + y = 0

So let us assume the general solution y = x^m

Then y’ = m x ^m-1

And Y’’ = m(m-1) x ^m-2

Now substitute in given ivp x²y’’ - xy’ + y = 0

x²(m(m-1) x ^m-2) - x(mx ^m-1) + x^m = 0

x²(m(m-1) x ^m x^-2) - xmx^mx^-1 + x^m = 0

m(m-1) x ^m - mx^m + x^m = 0

x^m ([m(m-1) - m + 1])= 0

but x^m is not equal to Zero so [m(m-1) - m + 1]= 0

m²-m - m + 1= 0 => m(m-1) – (m-1)= 0 => (m-1)(m-1) = 0

so m1 = m2 = 1

So when the roots are real and repeat then the general solution is in the form of

Y(x) = C₁x^m1 + C₂ x^m2 (lnx)

Y(x) = C₁x + C₂ x (lnx)

Now we have initial condition y(1) = -5

Y(x) = C₁x + C₂ x (lnx)

Y(1) = C₁(1) + C₂ (1) (ln(1))

-5 = C₁(1) + C₂ (1) (0)

-5 = C₁

Now we have y’(1) = 0

Y(x) = C₁x + C₂ x (lnx)

Y’(x) = C₁ + C₂ (lnx) + C₂ x (1/x)

Y’(x) = C₁ + C₂ (lnx) + C₂

Y’(1) = C₁ + C₂ (ln(1)) + C₂

0 = C₁ + C₂ (0) + C₂

0 = C₁ + C₂ => C₂ = -C₁

And C₁ = -5 then C₂ = -(-5) => C₂ = 5

Now the general solution

Y(x) = -5x + 5 x (lnx)

Y(x) = ((lnx) – 1) 5x