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

Step-by-step Solution

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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

Divide all the terms of the differential equation by $x^3$

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

Simplify the fraction by $x$

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

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

$x^{0}\frac{dy}{dx}+\frac{3x^2y}{x^3}=\frac{1}{x^{2}}$

Any expression (except $0$ and $\infty$) to the power of $0$ is equal to $1$

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

Simplify the fraction by $x$

$\frac{3y}{x^{3-2}}$

Subtract the values $3$ and $-2$

$\frac{3y}{x^{1}}$

Any expression to the power of $1$ is equal to that same expression

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

Simplifying

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

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, $\displaystyle\int\frac{1}{x}dx=\ln(x)$

$3\ln\left|x\right|$
4

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

$x^3$
5

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 $x$

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

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

$\frac{dy}{dx}x^3+3yx^{2}=x$
7

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

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

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

$\int\frac{d}{dx}\left(x^3y\right)dx=\int xdx$
9

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 $C$

$\frac{1}{2}x^2+C_0$
10

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 $x^3$

$y=\frac{x^2+C_1}{2x^3}$
11

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

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

e
π
ln
log
log
lim
d/dx
Dx
|◻|
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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