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Solve the differential equation $\frac{dy}{dx}+\frac{3}{x}y=\frac{1}{x^2}$

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Solution

Final answer to the problem

$y=\frac{x^2+C_1}{2x^3}$
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Step-by-step Solution

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  • Linear Differential Equation
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1

Multiplying the fraction by $y$

$\frac{dy}{dx}+\frac{3y}{x}=\frac{1}{x^2}$
2

We can identify that the differential equation has the form: $\frac{dy}{dx} + P(x)\cdot y(x) = Q(x)$, so we can classify it as a linear first order differential equation, where $P(x)=\frac{3}{x}$ and $Q(x)=\frac{1}{x^2}$. In order to solve the differential equation, the first step is to find the integrating factor $\mu(x)$

$\displaystyle\mu\left(x\right)=e^{\int P(x)dx}$

Compute the integral

$\int\frac{3}{x}dx$

The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$

$3\ln\left|x\right|$
3

To find $\mu(x)$, we first need to calculate $\int P(x)dx$

$\int P(x)dx=\int\frac{3}{x}dx=3\ln\left(x\right)$

Simplify $e^{3\ln\left|x\right|}$ by applying the properties of exponents and logarithms

$x^3$
4

So the integrating factor $\mu(x)$ is

$\mu(x)=x^3$

Multiplying the fraction by $x^3$

$\frac{dy}{dx}x^3+\frac{3yx^3}{x}=\frac{1}{x^2}x^3$

Multiplying the fraction by $x^3$

$\frac{dy}{dx}x^3+\frac{3yx^3}{x}=\frac{1x^3}{x^2}$

Any expression multiplied by $1$ is equal to itself

$\frac{dy}{dx}x^3+\frac{3yx^3}{x}=\frac{x^3}{x^2}$

Simplify the fraction $\frac{x^3}{x^2}$ by $x$

$\frac{dy}{dx}x^3+\frac{3yx^3}{x}=x$

Simplify the fraction $\frac{3yx^3}{x}$ by $x$

$\frac{dy}{dx}x^3+3yx^{2}=x$
5

Now, multiply all the terms in the differential equation by the integrating factor $\mu(x)$ and check if we can simplify

$\frac{dy}{dx}x^3+3yx^{2}=x$
6

We can recognize that the left side of the differential equation consists of the derivative of the product of $\mu(x)\cdot y(x)$

$\frac{d}{dx}\left(x^3y\right)=x$
7

Integrate both sides of the differential equation with respect to $dx$

$\int\frac{d}{dx}\left(x^3y\right)dx=\int xdx$
8

Simplify the left side of the differential equation

$x^3y=\int xdx$

Applying the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, in this case $n=1$

$\frac{1}{2}x^2$

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$

$\frac{1}{2}x^2+C_0$
9

Solve the integral $\int xdx$ and replace the result in the differential equation

$x^3y=\frac{1}{2}x^2+C_0$

Multiplying the fraction by $x^2$

$x^3y=\frac{x^2}{2}+C_0$

Combine all terms into a single fraction with $2$ as common denominator

$x^3y=\frac{x^2+2\cdot C_0}{2}$

We can rename $2\cdot C_0$ as other constant

$x^3y=\frac{x^2+C_1}{2}$

Divide both sides of the equation by $x^3$

$y=\frac{x^2+C_1}{2x^3}$
10

Find the explicit solution to the differential equation. We need to isolate the variable $y$

$y=\frac{x^2+C_1}{2x^3}$

Final answer to the problem

$y=\frac{x^2+C_1}{2x^3}$

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Function Plot

Plotting: $\frac{dy}{dx}+\frac{3}{x}y+\frac{-1}{x^2}$

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a
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c
d
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g
m
n
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v
w
x
y
z
.
(◻)
+
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×
◻/◻
/
÷
◻2
◻◻
√◻
√

◻√◻
◻√
∞
e
π
ln
log
log◻
lim
d/dx
D□x
∫
∫◻
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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