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Un Sol - Un Sol ( a.k.a. 1 + 1 = 2 Enamorados) is the first album released by Mexican singer Luis Miguel. Released in 1982, Miguel was twelve years old. es:1+1=2 enamorados..
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Particular solution of a differential equation
A solution obtained, by assigning particular values to the arbitrary constants in the general solution of the differential equation, is called its particular solution..
Which of the following is the particular solution for the given differ..
Which of the following is the particular solution for the given differential equation d y d x = x ² y such that y = 3 when x = 0? => y 2 = ( 2 / 3 ) x 3 + 9 or y 2 = - 2 3 x ³ + 9 or y 2 = x 3 + 9 or y 2 = 2 3 x ³ - 18..
Which of the following is the particular solution for the given differ..
Which of the following is the particular solution for the given differential equation d y d x ( x ³ + 28) = x ² y such that y 2 = 6 when x = - 3? => y ² = ( 2 ln | x ³ - 3 8 | ) 3 + 7 or y ² = ( 2 ln | x ³ + 3 8 | ) 3 + 6 or y &s..
Which of the following is the particular solution for the given differ..
Which of the following is the particular solution for the given differential equation (5 - 2 x ) y = d y d x , given that y = 2 when x = 0? => y = 4 e 5 x - x 2 or y = - 2 e 5 x + x 2 or | y | = 2 e 5 x - x 2 or y = e..
Particular Solutions to Differential EquationsLearn how to make particular solutions go to differential equations.
A Solution of Euler's Type for an Exact Differential Equation(dx, dy) = 0 ). So the contours of the Mathematica built-in function contourplot are particular solutions of the equation. To find ... Contributed by: Izidor Hafner
Question : Find the general solution to the differential equation dy/dx= 2xy/x^2 +1
Find the particular solution to the differential equation: 4e^y= y' - 3xe^y, y(0)=0
Show work please.
Answer : Your first equation is: dy/dx = 2*x*y/(x^2) + 1 Cancel the factor of x in the first term on the right hand side, and write this into 'standard form': dy/dx - (2/x)*y = 1 This is a first-order, linear, ordinary differential equation. There are several ways to solve this equation, but using an integrating factor will always work for such equations. For an equation of the form dy/dx + a(x)*y = b(x) we define an integrating factor as: p(x) = exp(INTEGRAL of {a(x) dx}) The solution is then given by: y(x) = (1/p(x))*INTEGRAL of { b(x) * p(x) dx} Here, a(x) = -2/x, and b(x) = 1, so: p(x) = exp(INTEGRAL of {-2/x dx}) p(x) = exp(-2ln(x)) = exp(ln((1/x^2))) = 1/x^2 The solution is given by: y(x) = (x^2)*INTEGRAL of {dx/x^2} y(x) = (x^2)*[-1/x + c] = c*x^2 - x where c is the constant of integration. ------------- Your second equation is: 4*exp(y) = dy/dx - 3x*exp(y) dy/dx = exp(y) * ( 4 + 3x) This is a separable, nonlinear, first-or..
Answer : Your first equation is: dy/dx = 2*x*y/(x^2) + 1 Cancel the factor of x in the first term on the right hand side, and write this into 'standard form': dy/dx - (2/x)*y = 1 This is a first-order, linear, ordinary differential equation. There are several ways to solve this equation, but using an integrating factor will always work for such equations. For an equation of the form dy/dx + a(x)*y = b(x) we define an integrating factor as: p(x) = exp(INTEGRAL of {a(x) dx}) The solution is then given by: y(x) = (1/p(x))*INTEGRAL of { b(x) * p(x) dx} Here, a(x) = -2/x, and b(x) = 1, so: p(x) = exp(INTEGRAL of {-2/x dx}) p(x) = exp(-2ln(x)) = exp(ln((1/x^2))) = 1/x^2 The solution is given by: y(x) = (x^2)*INTEGRAL of {dx/x^2} y(x) = (x^2)*[-1/x + c] = c*x^2 - x where c is the constant of integration. ------------- Your second equation is: 4*exp(y) = dy/dx - 3x*exp(y) dy/dx = exp(y) * ( 4 + 3x) This is a separable, nonlinear, first-or..
Question : What would you use for the particular solution of the following differential equations:
For example: y'=y .. you would use y=e^x + C
1. y'= -x/y
2. y' = x+y
3. y' = x(1+y)(2-y) *GENERAL EQUATION
Answer : The first equation is separable: dy/dx = -x/y y dy = -x dx (y^2)/2 = -(x^2)/2 + c where c is a constant of integration y^2 = -x^2 + d where I've absorbed the factor of 2 into the constant. y(x) = +/- sqrt(d - x^2) ----- The second equation is a first-order linear equation that is easily solved using an integrating factor (IF): dy/dx - y = x IF = exp(INTEGRAL of {-1 dx}) = exp(-x) The solution is given by: y(x) = exp(x)*INTEGRAL of {x*exp(-x) dx} y(x) = exp(x)* [c - exp(-x) - x*exp(-x)] y(x) = c*exp(x) - x - 1 --------- The third is also separable: dy/dx = x*(1+y)*(2-y) dy/(1+y)*(2-y) = x dx Expand the left hand side using partial fractions: (1/3)*(1/(1+y) - 1/(y-2)) dy = x dx 1/3*(ln(1+y) - ln(y-2)) = (x^2)/2 + c ln((y+1)/(y-2)) = (3/2)*x^2 + d (y+1)/(y-2) = exp((3x^2)/2 + d) = exp(d) * exp((3x^2)/2) = a*exp((3x^2)/2) where a is exp(d) is just another way of writing the constant of integration. This is an imp..
Answer : The first equation is separable: dy/dx = -x/y y dy = -x dx (y^2)/2 = -(x^2)/2 + c where c is a constant of integration y^2 = -x^2 + d where I've absorbed the factor of 2 into the constant. y(x) = +/- sqrt(d - x^2) ----- The second equation is a first-order linear equation that is easily solved using an integrating factor (IF): dy/dx - y = x IF = exp(INTEGRAL of {-1 dx}) = exp(-x) The solution is given by: y(x) = exp(x)*INTEGRAL of {x*exp(-x) dx} y(x) = exp(x)* [c - exp(-x) - x*exp(-x)] y(x) = c*exp(x) - x - 1 --------- The third is also separable: dy/dx = x*(1+y)*(2-y) dy/(1+y)*(2-y) = x dx Expand the left hand side using partial fractions: (1/3)*(1/(1+y) - 1/(y-2)) dy = x dx 1/3*(ln(1+y) - ln(y-2)) = (x^2)/2 + c ln((y+1)/(y-2)) = (3/2)*x^2 + d (y+1)/(y-2) = exp((3x^2)/2 + d) = exp(d) * exp((3x^2)/2) = a*exp((3x^2)/2) where a is exp(d) is just another way of writing the constant of integration. This is an imp..