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ydy+(xy2-8x)dx=0,y=3 when x=1.

Short Answer

Expert verified

Answer:

The solution containing one arbitrary constant is Iny2-8+x2=C, and with boundary condition, the value of the constant is C = 0. The particular solution is Iny2-8+22=1, and the plot of a slope field and some solution curves are as follows.

Step by step solution

01

Given Information

The given differential equation isydy+xy2-8xdx=0 with boundaries y = 3 when x = 1.

02

Definition of Differential Equation

A differential equation is a mathematical equation that connects one or more unknown functions with their derivatives. The study of differential equations primarily entails looking at their solutions (the set of functions that satisfy each equation) and characteristics.

03

Separate the variables

Separate the variables in ydy+xy2-8xdx=0 and keep y on one side and x on the other.

ydy+xy2-8xdx=0ydy=-xy2-8dxyy2-8dy=-xdx

04

Integrate the differential equation

Integrate the above differential equation using the variable separable form.

Substitute,

Solve the separated form of differential equation with an arbitrary constant.

Re-substitutein the above equation.

…… (1)

05

Find the value of the arbitrary constant and particular solutions

Put the boundary conditions,when, in (1) and find the value of.

Substitute the value ofin (1) and find the particular solution.

06

Plot the slope field

Plot the slope field of and some solution curves.

Therefore, for the differential equation , the solution containing one arbitrary constant is , and with boundary condition, when . ,The value of constant is , and the particular solution is ., The plot of a slope field and some of solution curves are as follows.,

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Most popular questions from this chapter

(a)Consider a light beam travelling downward into the ocean. As the beam progresses, it is partially absorbed and its intensity decreases. The rate at which the intensity is decreasing with depth at any point is proportional to the intensity at that depth. The proportionality constant μis called the linear absorption coefficient. Show that if the intensity at the surface is I0, the intensity at a distance s below the surface is I=I0eμs. The linear absorption coefficient for water is of the order of 10.2ft.1(the exact value depending on the wavelength of the light and the impurities in the water). For this value of μ, find the intensity as a fraction of the surface intensity at a depth of 1 ft, ft,ft,mile. When the intensity of a light beam has been reduced to half its surface intensity (I=12I0), the distance the light has penetrated into the absorbing substance is called the half-value thickness of the substance. Find the half-value thickness in terms of μ. Find the half-value thickness for water for the value of μgiven above.

(b) Note that the differential equation and its solution in this problem are mathematically the same as those in Example 1, although the physical problem and the terminology are different. In discussing radioactive decay, we call λthe decay constant, and we define the half-life T of a radioactive substance as the time when N=12N0(compare half-value thickness). Find the relation between λand T.

By using Laplace transforms, solve the following differential equations subject to the given initial conditions.

y"-4y'+4y=4, ∑0=0,ψ0=-2

By using Laplace transforms, solve the following differential equations subject to the given initial conditions.

y"-4y'=-4te2t, y0=0,y0'=1

By using Laplace transforms, solve the following differential equations subject to the given initial conditions.

y"+y=sint,

By using Laplace transforms, solve the following differential equations subject to the given initial conditions.

y¨+2y˙+5y=10cost,y0=2,y˙0=1

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