to 4x^2y"+17y=0, y(1)=-1, y'(1)= -1/2&nbsp;

Question

to 4x^2y&quot;+17y=0, y(1)=-1, y&#039;(1)= -1/2&amp;nbsp;

Vasquez · Accepted Answer

To solve the given differential equation:

4 x^{2} y + 17 y = 0

We'll use the method of integrating factors. First, we need to put the equation into standard form:

\frac{d y}{d x} + \frac{17}{4 x^{2}} y = 0

Here,

P (x) = \frac{17}{4 x^{2}}

and

Q (x) = 0

.
Next, we'll find the integrating factor

I (x)

, which is given by:

I (x) = e^{\int P (x) d x}

So, we'll first find

\int P (x) d x

:

\int \frac{17}{4 x^{2}} d x = - \frac{17}{4 x} + C

where

C

is the constant of integration. Therefore,

I (x) = e^{- \frac{17}{4 x}}

Multiplying both sides of the differential equation by

I (x)

, we get:

e^{- \frac{17}{4 x}} \frac{d y}{d x} + \frac{17}{4 x^{2}} e^{- \frac{17}{4 x}} y = 0

Recognizing that the left-hand side is the product rule of

(e^{- \frac{17}{4 x}} y)^{'}

, we can simplify the equation to:

(e^{- \frac{17}{4 x}} y)^{'} = 0

Integrating both sides with respect to

x

, we get:

e^{- \frac{17}{4 x}} y = C_{1}

where

C_{1}

is the constant of integration. Solving for

y

, we get:

y (x) = C_{1} e^{\frac{17}{4 x}}

Using the initial condition

y (1) = - 1

, we get:

- 1 = C_{1} e^{\frac{17}{4}}

Solving for

C_{1}

, we get:

C_{1} = - \frac{1}{e^{\frac{17}{4}}}

Therefore, the particular solution to the differential equation is:

y (x) = - \frac{1}{e^{\frac{17}{4}}} e^{\frac{17}{4 x}} = - \frac{1}{e^{(\ln e^{\frac{17}{4}}) / 4 x}} = - e^{- \frac{17}{4 x}}

Finally, we can use the initial condition

y^{'} (1) = - \frac{1}{2}

to solve for the constant of integration

C_{2}

:

y^{'} (x) = \frac{17}{4} e^{- \frac{17}{4 x}} x^{- 2}

So,

y^{'} (1) = - \frac{17}{4}

, which gives us:

- \frac{17}{4} = C_{2}

Therefore, the complete solution to the differential equation is:

y (x) = - e^{- \frac{17}{4 x}} - \frac{17}{4}

to 4x^2y"+17y=0, y(1)=-1, y'(1)= -1/2