Expert answers to "Find the general solution to the equation x(dydx)+3(y+x2)=sin⁡xx"

Name: Find the general solution to the equation x(dydx)+3(y+x2)=sin⁡xx
Uploaded: 2022-02-23T17:22:57+00:00
Description: Find the general solution to the equation x(dydx)+3(y+x2)=sin⁡xx

UkusakazaL

Answered question

2021-02-20

Find the general solution to the equation $x (\frac{d y}{d x}) + 3 (y + x^{2}) = \frac{\sin x}{x}$

Answer & Explanation

Lacey-May Snyder

Skilled2021-02-21Added 88 answers

Write this equation as follows:

\frac{d y}{d x} + 3 \frac{y}{x} + \frac{\sin x}{x^{2}} - 3 x (1)

We first solve the differential equation

\frac{d y}{d x} + 3 \frac{y}{x} = 0 \Rightarrow \frac{d y}{y} = - 3 \frac{d x}{x} \Rightarrow \int \frac{d y}{y} = - 3 \int \frac{d x}{x}

Thus,

\ln | y | = - 3 \ln | x | + C = \frac{\ln 1}{| x^{3} |} + C

Here we used that

α \ln β = I n β^{α}

. Now use the exponential function:

| y | = \frac{1}{| x^{3} |} \cdot e^{C}

Recall that

| α | = \pm α

, so we can write

y = \pm e^{C} \cdot \frac{1}{x^{3}}

Defining a new constant

D = \pm e^{C}

yields

y = \frac{D}{x^{3}}

This is the solution of the homogeneous equation. To obtain the solution of the initial equation, we regard D as a function of x, so

\frac{d y}{d x} = \frac{D^{'} x^{3} - 3 D x^{2}}{x^{6}} = \frac{D^{'}}{x^{3}} - 3 \frac{D}{x^{4}}

Plugging this and (2) into (1) yields

\frac{D^{'}}{x^{3}} - 3 \frac{D}{x^{4}} + 3 \frac{D}{x^{4}} = \frac{\sin x}{x^{2}} - 3 x \Rightarrow \frac{D^{'}}{x^{3}} = \frac{\sin x}{x^{2}} - 3 x

Thus,

D^{'} = x \sin x - 3 x^{4} \Rightarrow D = \int (x \sin x - 3 x^{4}) d x = \int x \sin x d x - 3 \int x^{4} d x

Now,

\int x \sin x d x = {(u = x \Rightarrow d u = d x), (d v = \sin x d x \Rightarrow v = - \cos x)}

= - x \cos x + \int \cos x d x

= - x \cos x + \sin x + C

Therefore,

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