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Find the integral surface of quasi linear partial differential equation xzp+yzq=-xy which passes through the curve y=x²,z=x³



Solve -y''-2y'+2y = e-2x

y(0)=1

y(infinity )=0


Solve yz2 (x2-yz)dx+zx2 (y2-zx)dy+xy2 (z2-xy)dz=0


1/xy2 +y4 is an integrating factor for the differential equation (x2y + y2)dx + (y3 - x3)dy = 0

x1/


Find the general solutions of the following system

  1. dx/dt = -4x+2y , dy/dt = 5x/2 + 2y

2. X' = ( -1 3

- 3 5) X

3. X' = ( 4 -5

5 -4)X



Differentiate with respect to the variable


(1) y=5/x^4 find dy/dx


(2) y=sin3t, find dy/dt


(3) 11sin4t - 5cos5t, find dv/dt



(all workings out)


Using the method of undetermined coefficients, find the general solution of the differential equation y^(iv) -2y^''' +y^''=3 e^-x+2 e^x x + e^-x sin (x ) 


y=sin3t find dy/dt


v=11sin4t - 5cos5t, find dv/dt



differentiate with respect to x


y=12e5x


y=ln3x


(D3 − 3D2 − 6D + 8)y = xe−3x


A bike is accelerating in yz-plane with its speed given by ()


at


t(w2) +2 * u2: dvt)


at


= u2


= u3 * t(2+w3)+3*u3, subject to the initial conditions,


v,(0) = n2; v0) = n3. Determine its speed at later time


u2 = 4 + 0.3R, u3 = 2+ 0.4R, w2 = 0.5+ R,w3 = 1+ R, n2 = 3.1+ 0.2R, n3 = 4.1 + 0.1R, p1 = 5.4 + 0.2R


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