Determine whether each first-order differntial equation is separable, linear, bo

pedzenekO

pedzenekO

Answered question

2021-09-20

Check to see if each first-order differential equation is linear, separable, both, or neither:
a)  dy  dx +exy=x2y2
b) y+sinx=x3y
c) lnxx2y=xy
d)  dy  dx +cosy=tanx

Answer & Explanation

Talisha

Talisha

Skilled2021-09-21Added 93 answers

Step 1
A first-order differential equation is said to be separable if it can be written as
dydx=f(x)g(y), f and g are known functions.
A first-order differntial equation is said to be separable if it can be written as
dydx+p(x)y=q(x), p(x) and q(x) are continuouse functions.
a) Consider the differential equation dydx+exy=x2y2
Rewrite the given differential equation as follows:
dydx=x2y2exy
=y(x2yex)
As the variables x and y cannot be separated and written as separate functions of x and y, given equation is not separable.
Also, in the equation dydx+exy=x2y2, p(x)=ex but right hand side is not a function of x alone. Hence, it is not a linear equation.
Therefore, the given differential equation is neither a separable equation nor a linear equation.
b) Consider the differential equation y+sinx=x3y
Rewrite the given differential equation as follows:
y+sinx=x3y
y+sinx=x3dydx
yx3+sinxx3=dydx
sinxx3dydxyx3
Thus, the given equation can be written as dydx+(1x3)y=sinxx3 where
p(x)=1x3 and q(x)=sinxx3
Hence, it is a linear equation.
Also, equation can be written as dy
alenahelenash

alenahelenash

Expert2023-05-13Added 556 answers

Step 1:
a) dydx+exy=x2y2
To determine if the equation is linear or separable, we need to rewrite it in a standard form. In this case, we observe that the equation is neither linear nor separable, as the terms involving y and y cannot be separated.
Step 2:
b) y+sinx=x3y
To check linearity, we need to express the equation in the form y+P(x)y=Q(x). We can rewrite the given equation as:
y1x3y=sinxx3
The equation is linear, as it can be expressed in the desired form.
Step 3:
c) lnxx2y=xy
To determine if the equation is linear or separable, we need to rewrite it in a standard form. Rearranging the terms, we have:
xyx2y=lnx
The equation is linear since it can be written in the form y+P(x)y=Q(x).
Step 4:
d) dydx+cosy=tanx
To check linearity, we need to express the equation in the form y+P(x)y=Q(x). We can rewrite the given equation as:
dydx+cosytanx=0
xleb123

xleb123

Skilled2023-05-13Added 181 answers

Answer:
(a)equation neither linear nor separable
(b)equation is linear
(c)equation is separable
(d)equation is linear
a) dydx+exy=x2y2
To determine if this equation is linear or separable, we need to rewrite it in a standard form. Rearranging the terms, we get:
dydx=x2y2exy
This equation is neither linear nor separable since it cannot be expressed in the form dydx=f(x)g(y) or dydx=ax+by+c.
b) y+sinx=x3y
Rearranging the terms, we have:
x3yy=sinx
This equation is linear since it can be expressed in the form y+P(x)y=Q(x). Here, P(x)=1x3 and Q(x)=sinx.
c) lnxx2y=xy
Rearranging the terms, we obtain:
xyx2y=lnx
This equation is separable since we can rewrite it in the form dydx=lnxx+xy.
d) dydx+cosy=tanx
This equation is linear since it can be expressed in the form dydx+P(x)y=Q(x). Here, P(x)=cosy and Q(x)=tanx.
star233

star233

Skilled2023-05-13Added 403 answers

a) dydx+exy=x2y2
To determine if this equation is linear or separable, we need to check its form. The equation is neither linear nor separable because the term exy appears. Therefore, the equation is neither linear nor separable.
b) y+sinx=x3y
Rearranging the equation, we have x3yy=sinx. To check if it's linear or separable, we need to see if we can isolate y or y on one side. Since y appears with x3, the equation is not separable. However, we can rewrite it as:
yyx3=sinxx3
Now we can see that the equation is linear. Applying an integrating factor method, the solution is:
y(x)=x3ex3dx(sinxx3)dx+Cx3
c) lnxx2y=xy
To determine if it's linear or separable, we can rearrange the equation as xyy=lnxx2y. Now we see that it's linear. Applying an integrating factor method, the solution is:
y(x)=xexdx(lnxx2)dx+Cx
d) dydx+cosy=tanx
Rearranging the equation, we have dydx=tanxcosy. This equation is neither linear nor separable because both y and x appear together in the trigonometric terms. Therefore, the equation is neither linear nor separable.

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