Question

If
y=ux

then dy/dx=(u+x du/dx)

i need explination how we got this??

Answer #1

A Bernoulli differential equation is one of the form
dy/dx+P(x)y=Q(x)y^n (∗)
Observe that, if n=0 or 1, the Bernoulli equation is linear. For
other values of n, the substitution u=y^(1−n) transforms the
Bernoulli equation into the linear equation
du/dx+(1−n)P(x)u=(1−n)Q(x).
Consider the initial value problem xy′+y=−8xy^2, y(1)=−1.
(a) This differential equation can be written in the form (∗)
with P(x)=_____, Q(x)=_____, and n=_____.
(b) The substitution u=_____ will transform it into the linear
equation du/dx+______u=_____.
(c) Using the substitution in part...

Solve the differential equation:
x*(du(x)/dx)=[x-u(x)]^2+u(x)
Hint: Let y=x-u(x)
Answer: u(x)=x-x/(x-K) with K=constant

To compute the integral ∫cos3(x)esin(x)dx, we should use first
the substitution u= __________ to obtain the integral ∫F(u)du,
where F(u)= __________ To compute the integral ∫F(u)du, we use the
method of integration by parts with
f(u)= __________ and g′(u)= __________ to get
∫F(u)du=G(u)−∫H(u)du , where G(u)= __________ and H(u)= __________
Now, to compute the integral ∫H(u)du, we need to use the method of
integration by parts a second time with f(u)= __________ and g′(u)=
__________ to get
∫H(u)du= __________ +C...

Solve the differential equation: dy/dx - y =e^x*y^2 (Using
u=y^-1)

what is the orthagonal form of the following equation: x(du/dx)
+ u = (2u/(u^2 +1)) ?

(1 point) Given the following differential equation
(x2+2y2)dxdy=1xy,
(a) The coefficient functions are M(x,y)= and N(x,y)= (Please input
values for both boxes.)
(b) The separable equation, using a substitution of y=ux, is
dx+ du=0 (Separate the variables with x with dx only and u with du
only.) (Please input values for both boxes.)
(c) The solution, given that y(1)=3, is
x=
Note: You can earn partial credit on this
problem.
I just need part C. thank you

Homogenous Differential Equations:
dy/dx = y - 4x / x-y
dy/dx = - (4x +3y / 2x+y)

Homogeneous Differential Equations:
dy/dx = xy/x^(2) - y^(2)
dy/dx = x^2 + y^2 / 2xy

Solve the given differential equation
y-x(dy/dx)=3-2x2(dy/dx)

Solve: 1.dy/dx=(e^(y-x)).secy.(1+x^2),y(0)=0.
2.dy/dx=(1-x-y)/(x+y),y(0)=2 .

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