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Keith_Richards [23]
4 years ago
3

It can be helpful to classify a differential equation, so that we can predict the techniques that might help us to find a functi

on which solves the equation. Two classifications are the order of the equation -- (what is the highest number of derivatives involved) and whether or not the equation is linear .
Linearity is important because the structure of the the family of solutions to a linear equation is fairly simple. Linear equations can usually be solved completely and explicitly. Determine whether or not each equation is linear:
1. (1+y2)(d2y/dt2)+t(dy/dt)+y=et

2. t2(d2y/dt2)+t(dy/dt)+2y=sin t

3. (d3y/dt3)+t(dy/dt)+(cos2(t))y=t3

4. y''-y+y2=0

You have 10 choices to choose for each problem:

a. 1st. order linear differential equation

b. 2nd. order linear differential equation

c. 3rd. order linear differential equation

d. 4th. order linear differential equation

e. 5th. order linear differential equation

f. 1st. order non-linear differential equation

g. 2nd. order non-linear differential equation

h. 3rd. order non-linear differential equation

i. 4th. order non-linear differential equation

f. 5th. order non-linear differential equation
Mathematics
1 answer:
IRINA_888 [86]4 years ago
5 0
Answer:

1. g. 2nd. order non-linear differential equation

2. a. 1st. order linear differential equation

3. c. 3rd. order linear differential equation

4. g. 2nd. order non-linear differential equation

Step-by-step explanation:

QUESTION 1

(1+y^2)(\frac{d^2y}{dt^2})+t\frac{dy}{dt} +y=e^t

Order: The highest derivative present in this differential equation is the second derivative (\frac{d^2y}{dt^2}). Hence the order of this differential equation is 2.

Linearity: There is the presence of the product of the dependent variable , y and its derivative [(1+y^2)(\frac{d^2y}{dt^2})].  

Hence this differential equation is non-linear.

Classification: Second order non-linear ordinary differential equation.

QUESTION 2

The given differential equation is  

t^2\frac{d^2y}{dt^2}+t\frac{dy}{dt} +2y=\sin t

Order: The highest derivative present in this differential equation is \frac{d^2y}{dt^2}

Hence it is a second order differential equation.

Linearity: There is no presence of the product of the dependent variable and/or its derivative. There is no presence of higher powers of the dependent variable or its derivative. There is no transcendental function of the dependent variable.

Classification: First order linear differential equation

QUESTION 3

The given differential equation is

\frac{d^3y}{dt^3}+t\frac{dy}{dx} +\cos (2t)y=t^3

Order: The highest derivative present in this differential equation is \frac{d^3y}{dt^3}

Hence it is a third order differential equation.

Linearity: There is no presence of the product of the dependent variable and/or its derivative. There is no presence of higher powers of the dependent variable or its derivative. There is no transcendental function of the dependent variable.

Classification: Third order linear ordinary differential equation

QUESTION 4

The given differential equation is;

y"-y+y^2=0

Order: The highest derivative present in this differential equation is y"

Hence it is a second order differential equation.

Linearity: There is the presence of higher power of the dependent variable, y^2. Hence the differential equation is non-linear.

Classification: Second order non-linear differential equation
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If j(x)=3x+1, find x if j(x)=-11​
vladimir1956 [14]

Answer:

x = -4

Step-by-step explanation:

Step 1: Define

j(x) = 3x + 1

j(x) = -11

Step 2: Substitute and solve for <em>x</em>

-11 = 3x + 1

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3 years ago
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5 0
3 years ago
68. Solve: 46x - 10) = 8x + 40<br>A 0<br>B.5/2<br>ina<br>c. 23<br>D. 5​
Artist 52 [7]

Solve: 4(6x - 10) = 8x + 40

A 0

B.5/2

c. 23

D. 5​

<h3><u>Answer:</u></h3>

Option D

The solution to given equation is x = 5

<h3><u>Solution:</u></h3>

Given that we have to solve the given equation

4(6x - 10) = 8x + 40

Let us solve the above expression and find value of "x"

Multiplying 4 with terms inside bracket in L.H.S we get,

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4 years ago
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