Mister Exam

# Differential equation xy'-3y=6

Which of the following is a solution to the differential equation xy′−3y=6?

y() =
y'() =
y''() =
y'''() =
y''''() =

from to

### The solution

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            d
-3*y(x) + x*--(y(x)) = 6
dx          
$$x \frac{d}{d x} y{\left(x \right)} - 3 y{\left(x \right)} = 6$$
x*y' - 3*y = 6
Detail solution
Divide both sides of the equation by the multiplier of the derivative of y':
$$x$$
We get the equation:
$$\frac{x \frac{d}{d x} y{\left(x \right)} - 3 y{\left(x \right)}}{x} = \frac{6}{x}$$
This differential equation has the form:
y' + P(x)y = Q(x)

where
$$P{\left(x \right)} = - \frac{3}{x}$$
and
$$Q{\left(x \right)} = \frac{6}{x}$$
and it is called linear inhomogeneous
differential first-order equation:
First of all, we should solve the correspondent linear homogeneous equation
y' + P(x)y = 0

with multiple variables
The equation is solved using following steps:
From y' + P(x)y = 0 you get

$$\frac{dy}{y} = - P{\left(x \right)} dx$$, if y is not equal to 0
$$\int \frac{1}{y}\, dy = - \int P{\left(x \right)}\, dx$$
$$\log{\left(\left|{y}\right| \right)} = - \int P{\left(x \right)}\, dx$$
Or,
$$\left|{y}\right| = e^{- \int P{\left(x \right)}\, dx}$$
Therefore,
$$y_{1} = e^{- \int P{\left(x \right)}\, dx}$$
$$y_{2} = - e^{- \int P{\left(x \right)}\, dx}$$
The expression indicates that it is necessary to find the integral:
$$\int P{\left(x \right)}\, dx$$
Because
$$P{\left(x \right)} = - \frac{3}{x}$$, then
$$\int P{\left(x \right)}\, dx$$ =
= $$\int \left(- \frac{3}{x}\right)\, dx = - 3 \log{\left(x \right)} + Const$$
Detailed solution of the integral
So, solution of the homogeneous linear equation:
$$y_{1} = x^{3} e^{C_{1}}$$
$$y_{2} = - x^{3} e^{C_{2}}$$
that leads to the correspondent solution
for any constant C, not equal to zero:
$$y = C x^{3}$$
We get a solution for the correspondent homogeneous equation
Now we should solve the inhomogeneous equation
y' + P(x)y = Q(x)

Use variation of parameters method
Now, consider C a function of x

$$y = x^{3} C{\left(x \right)}$$
And apply it in the original equation.
Using the rules
- for product differentiation;
- of composite functions derivative,
we find that
$$\frac{d}{d x} C{\left(x \right)} = Q{\left(x \right)} e^{\int P{\left(x \right)}\, dx}$$
Let use Q(x) and P(x) for this equation.
We get the first-order differential equation for C(x):
$$\frac{d}{d x} C{\left(x \right)} = \frac{6}{x^{4}}$$
So, C(x) =
$$\int \frac{6}{x^{4}}\, dx = Const - \frac{2}{x^{3}}$$
Detailed solution of the integral
use C(x) at
$$y = x^{3} C{\left(x \right)}$$
and we get a definitive solution for y(x):
$$x^{3} \left(Const - \frac{2}{x^{3}}\right)$$
                3
y(x) = -2 + C1*x 
$$y{\left(x \right)} = C_{1} x^{3} - 2$$
The classification
separable
1st exact
1st linear
Bernoulli
almost linear
linear coefficients
separable reduced
lie group
nth linear euler eq nonhomogeneous undetermined coefficients
nth linear euler eq nonhomogeneous variation of parameters
separable Integral
1st exact Integral
1st linear Integral
Bernoulli Integral
almost linear Integral
linear coefficients Integral
separable reduced Integral
nth linear euler eq nonhomogeneous variation of parameters Integral
The graph
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