Question

(1) obtain the solution for the initial value problem dy/dx+(cotx)y=xcscx,y(pi/2)=1

(2)solve the differential equation dy/dx=x(1+y^2)

(3) solve the differential equation dy/dx+(cotx)y=xcscx

Answer 1

Answer on Question #68322 - Math - Differential Equations

Question

(1) Obtain the solution for the initial value problem $\mathrm{dy} / \mathrm{dx} + (\mathrm{cotx})\mathrm{y} = \mathrm{xcsx},\mathrm{y}(\mathrm{pi} / 2) = 1$

(2) Solve the differential equation $\mathrm{dy} / \mathrm{dx} = \mathrm{x}(1 + \mathrm{y}^{\wedge}2)$

(3) Solve the differential equation $\mathrm{dy} / \mathrm{dx} + (\mathrm{cotx})\mathrm{y} = \mathrm{xcsx}$

Solution

Q1. We have the initial value problem

$$\frac {d y}{d x} + (\cot x) y = x \csc x \quad y \left(\frac {\pi}{2}\right) = 1$$

First we find the general solution of the equation, i.e. solve the problem

$$\frac {d y}{d x} + (\cot x) y = x \csc x$$

This equation is a linear equation of the form

$$\frac {d y}{d x} + P (x) y = Q (x)$$

where

$$P (x) = \cot x, Q (x) = x \csc x$$

To solve this differential equation, we need to multiply both sides by the integrating factor $I(x)$ , where

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

and integrate both sides.

Find integral

$$\int \cot x d x = \int \frac {\cos x}{\sin x} d x = \int \frac {d (\sin x)}{\sin x} = \ln | \sin x | + C$$

We are looking for a particular integrating factor, so we take $C = 0$ . Then we get

$$| I (x) | = e ^ {\ln | \sin x |} = | \sin x |$$

Multiply both sides of the equation by $I(x) = \sin x$

$$\sin x \frac {d y}{d x} + (\sin x \cdot \cot x) y = x (\csc x \cdot \sin x)$$

or

$$\sin x \frac {d y}{d x} + (\cos x) y = x$$

and so we can rewrite the equation as

$$\frac {d}{d x} (y \cdot \sin x) = x$$

Integrate both sides of this equation we get the general solution

$$\int \frac {d}{d x} (y \cdot \sin x) d x = \int x d x \Rightarrow y \cdot \sin x = \frac {x ^ {2}}{2} + c$$

where $c$ is constant defined by the initial conditions.

Now we find $c$ and solve the initial value problem. Substitute the initial condition

$$y \left(\frac {\pi}{2}\right) = 1$$

into the general solution:

$$1 \cdot \sin \left(\frac {\pi}{2}\right) = \frac {1}{2} \left(\frac {\pi}{2}\right) ^ {2} + c$$

or

$$1 = \frac {\pi^ {2}}{8} + c \Rightarrow c = 1 - \frac {\pi^ {2}}{8}$$

Then we obtain the solution for the initial value problem

$$y \cdot \sin x = \frac {x ^ {2}}{2} + 1 - \frac {\pi^ {2}}{8}$$

or dividing both sides by $\sin x$ ,

$$y = \left(\frac {x ^ {2}}{2} + 1 - \frac {\pi^ {2}}{8}\right) \csc x$$

Answer: the solution for the initial value problem is

$$y = \left(\frac {x ^ {2}}{2} + 1 - \frac {\pi^ {2}}{8}\right) \csc x$$

Q2. We have the equation

$$\frac {d y}{d x} = x (1 + y ^ {2})$$

This equation is the separable equation, i.e. it is a first-order differential equation in which the expression for $\frac{dy}{dx}$ can be written in the form

$$\frac {d y}{d x} = f (x) g (y)$$

For this problem

$$f (x) = x, \quad g (y) = (1 + y ^ {2})$$

To solve this equation we rewrite it in the differential form

$$d y = x (1 + y ^ {2}) d x$$

or, dividing both sides by $1 + y^{2}$

$$\frac {d y}{(1 + y ^ {2})} = x d x$$

Then we integrate both sides of this equation:

$$\int \frac {d y}{(1 + y ^ {2})} = \int x d x$$

We get

$$\arctan y = \frac {x ^ {2}}{2} + c$$

or

$$y = \tan \left(\frac {x ^ {2}}{2} + c\right),$$

where $c$ is constant

Answer: The solution of the given differential equation is

$$y = \tan \left(\frac {x ^ {2}}{2} + c\right)$$

Q3. We have the equation

$$\frac{dy}{dx} + (\cot x)y = x \csc x$$

The general solution of this equation we obtained by solving problem 1:

$$y \cdot \sin x = \frac{x^2}{2} + c$$

or dividing both sides by $\sin x$,

$$y = \left(\frac{x^2}{2} + c\right) \csc x$$

Answer: The solution of the given differential equation is

$$y = \left(\frac{x^2}{2} + c\right) \csc x$$

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