The special solution of dy / DX = e ^ (x-y-2), y (0) = 0

The special solution of dy / DX = e ^ (x-y-2), y (0) = 0


dy/dx=e^(x-y-2)
dy/dx=e^-2*e^x/e^-y
e^ydy=e^-2e^xdx
The integral on both sides is obtained
e^y=e^-2e^x+C
Substituting (0,0)
1=e^-2+C=0
e^y=e^-2e^x+1-e^-2



Given y ^ 2 (1 + x ^ 2) = C, find dy / DX


dy²(1+x²)=dc
y²d(1+x²)+(1+x²)dy²=0
2xy²dx+2y(1+x²)dy=0
So dy / DX = - XY / (1 + X & # 178;)



Y = x ^ 2 (√ x-sin2x), find dy


dy=[2x(√x-sin2x)+x^2(1/(2√x)-2cos2x]dx
Just take the lower derivative of Y and add DX



Y = sin2x, find dy


dy=cos2xd2x
=2cos2xdx

Elden Ring Premiere

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