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Anna007 [38]
3 years ago
8

Solve the given differential equation by an appropriate substitution.

Mathematics
1 answer:
ki77a [65]3 years ago
8 0

Answer:

y(x) = -x sqrt(c_1 x - 1) or y(x) = x sqrt(c_1 x - 1)

Step-by-step explanation:

Solve Bernoulli's equation ( dy(x))/( dx) = (x^2 + 3 y(x)^2)/(2 x y(x)):

Rewrite the equation:

( dy(x))/( dx) - (3 y(x))/(2 x) = x/(2 y(x))

Multiply both sides by 2 y(x):

2 ( dy(x))/( dx) y(x) - (3 y(x)^2)/x = x

Let v(x) = y(x)^2, which gives ( dv(x))/( dx) = 2 y(x) ( dy(x))/( dx):

( dv(x))/( dx) - (3 v(x))/x = x

Let μ(x) = e^( integral-3/x dx) = 1/x^3.

Multiply both sides by μ(x):

(( dv(x))/( dx))/x^3 - (3 v(x))/x^4 = 1/x^2

Substitute -3/x^4 = d/( dx)(1/x^3):

(( dv(x))/( dx))/x^3 + d/( dx)(1/x^3) v(x) = 1/x^2

Apply the reverse product rule f ( dg)/( dx) + g ( df)/( dx) = d/( dx)(f g) to the left-hand side:

d/( dx)(v(x)/x^3) = 1/x^2

Integrate both sides with respect to x:

integral d/( dx)(v(x)/x^3) dx = integral1/x^2 dx

Evaluate the integrals:

v(x)/x^3 = -1/x + c_1, where c_1 is an arbitrary constant.

Divide both sides by μ(x) = 1/x^3:

v(x) = x^2 (c_1 x - 1)

Solve for y(x) in v(x) = y(x)^2:

Answer: y(x) = -x sqrt(c_1 x - 1) or y(x) = x sqrt(c_1 x - 1)

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The fifth period class has a greater fraction of students that like to bowl.

Any number in the form of p/q where p and q are integers and q is not equal to 0 is a rational number. Examples of rational numbers are 1/2, -3/4, 0.3, or 3/10.

Here we have,

In second period, 37.5% students like to bowl

∴, 37.5% = 37.5/100

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Learn more about rational numbers by referring to the following link:

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