Question

Important Instructions: (1) λ is typed as lambda. (2) Use hyperbolic trig functions cosh(x) and sinh(x) instead of ex and e−x. (3) Write the functions alphabetically, so that if the solutions involve cos and sin, your answer would be Acos(x)+Bsin(x). (4) For polynomials use arbitrary constants in alphabetical order starting with highest power of x, for example, Ax2+Bx. (5) Write differential equations with leading term positive, so X′′−2X=0 rather than −X′′+2X=0. (6) Finally you need to simplify arbitrary constants. For example if A, B and C are arbitrary constants then AC, BC could be simplified to A, B since C in this case would be redundant. The PDE k2∂2u∂x2=y2∂u∂y is separable, so we look for solutions of the form u(x,t)=X(x)Y(y). When solving DE in X and Y use the constants A and B for X and C for Y. The PDE can be rewritten using this solution as (place constant k in the DE for Y) into X''/X = (1/k^2)(y^2)(Y'/Y) = −λ Note: Use the prime notation for derivatives, so the derivative of X and Y can be written as X′ and Y′. Do NOT use X′(x) Since these differential equations are independent of each other, they can be separated DE in X: X''+lambdaX =0 DE in Y: =0 (your answer should be simplified so there are no fractions) Now we solve the separated ODEs for the different cases in λ. In each case the general solution in X is written with constants A and B and the general solution in Y is written with constants C and D. Case 1: λ=0 X(x)= Ax+B Y(y)= C DE in Y If λ≠0, the differential equation in Y is first order, linear, and more importantly separable. We separate the two sides as Y'/Y = -((k^2)/(y^2))lambda Integrating both sides with respect to y (placing the constant of integration C in the right hand side) we get ln(absY) = Solving for Y, using the funny algebra of constants where eC=C is just another constant we get Y= For λ≠0 we get a Sturm-Louiville problem in X which we need to handle two more cases Case 2: λ=−a2 X(x)= Case 3: λ=a2 X(x)= Final Producet Solution (make sure to remove any redundant arbitrary constants) Case 1: λ=0 u= Ax+B Case 2: λ=−a2 u= Case 3: λ=a2 u=

Answer #1

Important Instructions: (1) λ is typed as lambda. (2) Use
hyperbolic trig functions cosh(x) and sinh(x) instead of ex and
e−x. (3) Write the functions alphabetically, so that if the
solutions involve cos and sin, your answer would be
Acos(x)+Bsin(x). (4) For polynomials use arbitrary constants in
alphabetical order starting with highest power of x, for example,
Ax2+Bx. (5) Write differential equations with leading term
positive, so X′′−2X=0 rather than −X′′+2X=0. (6) Finally you need
to simplify arbitrary constants. For...

Consider the first order separable equation
y′=12x^3y(1+2x^4)^1/2. An implicit general solution can be written
in the form y=Cf(x) for some function f(x) with C an arbitrary
constant. Here f(x)= Next find the explicit solution of the initial
value problem y(0)=1
y=

(A). Find the maximum of the following utility function with
respect to x;
U= x^2 * (120-4x).
The utility function is U(x,y)= sqrt(x) + sqrt(y) . The price of
good x is Px and the price of good y is Py. We denote income by M
with M > 0. This function is well-defined for x>0 and
y>0.
(B). Compute (aU/aX) and (a^2u/ax^2). Is the utility function
increasing in x? Is the utility function concave in x?
(C). Write down...

Consider the second order linear partial differential equation
a(x, y)uxx + 2b(x, y)uxy + c(x, y)uyy + d(x, y)ux + e(x, y)uy +
f(x, y)u = 0. (1)
a). Show that under a coordinate transformation ξ = ξ(x, y), η =
η(x, y) the PDE (1) transforms into the PDE α(ξ, η)uξξ + 2β(ξ,
η)uξη + γ(ξ, η)uηη + δ(ξ, η)uξ + (ξ, η)uη + ψ(ξ, η)u = 0, (2) where
α(ξ, η) = aξ2 x + 2bξxξy + cξ2...

(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

1 point)
If the series y(x)=∑n=0∞cnxny(x)=∑n=0∞cnxn is a solution of the
differential equation 1y″−6x2y′+2y=01y″−6x2y′+2y=0, then
cn+2=cn+2= cn−1+cn−1+ cn,n=1,2,…cn,n=1,2,…
A general solution of the same equation can be written as
y(x)=c0y1(x)+c1y2(x)y(x)=c0y1(x)+c1y2(x), where
y1(x)=1+∑n=2∞anxn,y1(x)=1+∑n=2∞anxn,
y2(x)=x+∑n=2∞bnxn,y2(x)=x+∑n=2∞bnxn,
Calculate
a2=a2= ,
a3=a3= ,
a4=a4= ,
b2=b2= ,
b3=b3= ,
b4=b4= .

(1 point) A Bernoulli differential equation is one of the
form
dydx+P(x)y=Q(x)yn (∗)
Observe that, if n=0 or 1, the Bernoulli equation is linear. For
other values of n, the substitution u=y1−n transforms the Bernoulli
equation into the linear equation
dudx+(1−n)P(x)u=(1−n)Q(x).dudx+(1−n)P(x)u=(1−n)Q(x).
Consider the initial value problem
y′=−y(1+9xy3), y(0)=−3.
(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
dudx+ u= .
(c) Using...

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...

Use the method for solving homogeneous equations to solve the
following differential equation.
(9x^2-y^2)dx+(xy-x^3y^-1)dy=0
solution is F(x,y)=C, Where C= ?

A) In this problem we consider an equation in differential form
???+???=0Mdx+Ndy=0. The equation
(5?^4?+4cos(2?)?^−4)??+(5?^5−3?^−4)??=0
in differential form ?˜??+?˜??=0 is not exact. Indeed, we
have
My-Mx= ________ 5x^4-4(4)cos(2x)*y^(-5)-25x^4
(My-Mn)/M=_____ -4/y
in function y alone.
?(?)= _____y^4
Multiplying the original equation by the integrating factor we
obtain a new equation ???+???=0 where
M=____
N=_____
which is exact since
My=_____
Nx=______
This problem is exact. Therefore an implicit general solution
can be written in the form ?(?,?)=? where
?(?,?)=_________
Finally find the value...

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