Question

differential equation one solution is given,

xy''-(2x+1)y'+(x+1)y=x^2; y_1=e^x

Answer #1

If you have any doubt then let me know in the comment.

solve differential equation
(x^2)y'' - xy' +y =2x

given the differential equation y'+y=x^2+2x, compute the value of
the solution y(x) at x=1 given the inital condition y(0)=2. Also,
use h=0.1

Given that y=e^x is a solution of the equation (x-1)y''-xy'+y=0,
find the general solution to (x-1)y''-xy'+y=1.

Given the second-order differential equation
y''(x) − xy'(x) + x^2 y(x) = 0
with initial conditions
y(0) = 0, y'(0) = 1.
(a) Write this equation as a system of 2 first order
differential equations.
(b) Approximate its solution by using the forward Euler
method.

Find the differential Equation of
xy'-2y+(2x^3)e^-x=0

Power series
Find the particular solution of the differential equation:
(x^2+1)y"+xy'-4y=0 given the boundary conditions x=0, y=1 and y'=1.
Use only the 7th degree term of the solution. Solve for y at x=2.
Write your answer in whole number.

Solve the Homogeneous differential equation
(7 y^2 + 1 xy)dx - 1 x^2 dy = 0
(a) A one-parameter family of solution of the equation is y(x)
=
(b) The particular solution of the equation subject to the
initial condition y(1) =1/7.

3. Find the general solution to the differential equation:
(x^2 + 1/( x + y) + y cos(xy)) dx + (y ^2 + 1 / (x + y) + x
cos(xy)) dy = 0

Find the solution of the Differential Equation
X^2y''-xy'+y=x

solve differential equation ((x)2 - xy +(y)2)dx - xydy
= 0
solve differential equation (x^2-xy+y^2)dx - xydy =
0

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