大学高等数学求解
如图
a(n)=n/2^n a(n+1)/a(n)=(n+1)2^n/[n2^(n+1)]--->1/2 ∴R=2
z=x^3y+xy^3 偏z/偏x=3x^2y+y^3 偏^2z/偏x^2=6xy∫
y'+p(x)y=0
dy/dx=x/y ydy=xdx y^2/2=x^2/2+C
∫[0,1]x(1-x)dx=∫[0,1](x-x^2)dx=[x^2/2-x^3/3]|[0,1]=1/6
∫[0,π/4]cos^2(x)dx=1/2∫[0,π/4][1-cos(2x)]dx=x/2|[0,π/4]-1/4sin(2x)|[0,π/4]=π/8-1/4
∫[0,π/4]xcosxdx=xsinx|[0,π/4]-∫[0,π/4]sinxdx=√2π/8+cosx|[0,π/4]=√2π/8+√2/2-1
∫[0,1]xe^xdx=xe^x|[0,1]-∫[0,1]e^xdx=e-e+1=1
∫[0,+∞]2/(1+x^2)dx=2arctanx|[0,+∞]=π
∫[0,1]dx/(1+e^x)=∫[0,1]e^(-x)dx/[1+e^(-x)]=-ln[1+e^(-x)]|[0,1]=ln2-ln(1+e)=ln[2/(1+e)]
z=x^y+y^2 dz=y*x^(y-1)dx+(x^ylnx+2y)dy
递减，一般项趋近于0，交错级数收敛
绝对收敛
sinx=x-x^3/3!+x^5/5!-x^7/7!+…… 逐项微分： cosx=1-x^2/2!+x^4/4!-x^6/6!+……
z=uv+50t u=e^t v=t^2 z'(t)=t^2*e^t+2t*e^t+50
yy''+2y'^2=0 设dy/dx=p(x) 则 y''=dp/dx=dp/dy*dy/dx=pdp/dy
ypdp/dy+2p^2=0 ypdp/dy=-2p^2 ydp/dy=-2p dp/p=-2dy/y
lnp=-2lny+lnC1 p=C1y^(-2)
dy/dx=C1y^(-2)
y^2dy=c1dx y^3/3=C1x+C2 通解：y^3=3C1x+3C2
