求微分方程[img=143x21]17da5f14490e50e.png[/img]的通解,实验命令为(). A: dsolve(D2y-2*Dy+5*y=sin(2*x),x)ans =exp(x)*sin(2*x)*C2+exp(x)*cos(2*x)*C1+1/17*sin(2*x)+4/17*cos(2*x) B: dsolve('D2y-2*Dy+5*y=sin(2*x)','x')ans =cos(2*x)*(sin(4*x)/17 - cos(4*x)/68 + 1/4) - sin(2*x)*(cos(4*x)/17 + sin(4*x)/68) + C1*cos(2*x)*exp(x) - C2*sin(2*x)*exp(x) C: dsolve(D2y-2*Dy+5*y=sin(2*x),'x','y')ans =exp(x)*sin(2*x)*C2+exp(x)*cos(2*x)*C1+1/17*sin(2*x)+4/17*cos(2*x)
求微分方程[img=143x21]17da5f14490e50e.png[/img]的通解,实验命令为(). A: dsolve(D2y-2*Dy+5*y=sin(2*x),x)ans =exp(x)*sin(2*x)*C2+exp(x)*cos(2*x)*C1+1/17*sin(2*x)+4/17*cos(2*x) B: dsolve('D2y-2*Dy+5*y=sin(2*x)','x')ans =cos(2*x)*(sin(4*x)/17 - cos(4*x)/68 + 1/4) - sin(2*x)*(cos(4*x)/17 + sin(4*x)/68) + C1*cos(2*x)*exp(x) - C2*sin(2*x)*exp(x) C: dsolve(D2y-2*Dy+5*y=sin(2*x),'x','y')ans =exp(x)*sin(2*x)*C2+exp(x)*cos(2*x)*C1+1/17*sin(2*x)+4/17*cos(2*x)
求微分方程[img=364x55]17da65386dfd612.png[/img]的通解; ( ) A: - cos(2*x)*exp(x)*(x/4 - sin(4*x)/16) + C23*cos(2*x)*exp(x) + C24*sin(2*x)*exp(x) B: (3*sin(2*x)*exp(x))/32 - (sin(6*x)*exp(x))/32 - cos(2*x)*exp(x)*(x/4 - sin(4*x)/16) + C23*cos(2*x)*exp(x) + C24*sin(2*x)*exp(x) C: - sin(4*x)/16) + C23*cos(2*x)*exp(x) + C24*sin(2*x)*exp(x) D: (sin(6*x)*exp(x))/32 - cos(2*x)*exp(x)*(x/4 - sin(4*x)/16) + C23*cos(2*x)*exp(x) + C24*sin(2*x)*exp(x)
求微分方程[img=364x55]17da65386dfd612.png[/img]的通解; ( ) A: - cos(2*x)*exp(x)*(x/4 - sin(4*x)/16) + C23*cos(2*x)*exp(x) + C24*sin(2*x)*exp(x) B: (3*sin(2*x)*exp(x))/32 - (sin(6*x)*exp(x))/32 - cos(2*x)*exp(x)*(x/4 - sin(4*x)/16) + C23*cos(2*x)*exp(x) + C24*sin(2*x)*exp(x) C: - sin(4*x)/16) + C23*cos(2*x)*exp(x) + C24*sin(2*x)*exp(x) D: (sin(6*x)*exp(x))/32 - cos(2*x)*exp(x)*(x/4 - sin(4*x)/16) + C23*cos(2*x)*exp(x) + C24*sin(2*x)*exp(x)
(1-sin^6α-cos^6α)/(sin^2α-sin^4α)的值
(1-sin^6α-cos^6α)/(sin^2α-sin^4α)的值
求不定积分[img=132x48]17da6537fc8dad6.png[/img]; ( ) A: -(4*(cos(x/2)/2 + 2*sin(x/2)))/(17*exp(2*x)) B: (4*(sin(x/2)/2 + 2*sin(x/2)))/(17*exp(2*x)) C: (4*(cos(x/2)/2 + 2*sin(x/2)))/(17*exp(2*x)) D: (4*(cos(x/2)/2 + 2*cos(x/2)))/(17*exp(2*x))
求不定积分[img=132x48]17da6537fc8dad6.png[/img]; ( ) A: -(4*(cos(x/2)/2 + 2*sin(x/2)))/(17*exp(2*x)) B: (4*(sin(x/2)/2 + 2*sin(x/2)))/(17*exp(2*x)) C: (4*(cos(x/2)/2 + 2*sin(x/2)))/(17*exp(2*x)) D: (4*(cos(x/2)/2 + 2*cos(x/2)))/(17*exp(2*x))
$\int {{{\sin 2x} \over {1 + {{\sin }^4}x}}} {\rm{d}}x = $ A: $\arctan (\sin x) + C$ B: $\arctan ({\sin ^2}x) + C$ C: ${\arctan ^2}(\sin x) + C$ D: $ - {\arctan ^2}(\sin x) + C$
$\int {{{\sin 2x} \over {1 + {{\sin }^4}x}}} {\rm{d}}x = $ A: $\arctan (\sin x) + C$ B: $\arctan ({\sin ^2}x) + C$ C: ${\arctan ^2}(\sin x) + C$ D: $ - {\arctan ^2}(\sin x) + C$
求下列不定积分.[tex=7.286x2.643]28VI4S//fW038PiMAbBHktfj3FfJYocy4+TgcP5gH+6DCjcL5MVe5w4GLCJx2oaC[/tex].腺 由于 $\sin ^{4} x+\cos ^{4} x=\left(\cos ^{2} x-\sin ^{2} x\right)^{2}+2 \sin ^{2} x \cos ^{2} x$$=\cos ^{2} 2 x+\frac{1}{2} \sin ^{2} 2 x$原式 $=\int \frac{\mathrm{d} x}{\cos ^{2} 2 x+\frac{1}{2} \sin ^{2} 2 x}$
求下列不定积分.[tex=7.286x2.643]28VI4S//fW038PiMAbBHktfj3FfJYocy4+TgcP5gH+6DCjcL5MVe5w4GLCJx2oaC[/tex].腺 由于 $\sin ^{4} x+\cos ^{4} x=\left(\cos ^{2} x-\sin ^{2} x\right)^{2}+2 \sin ^{2} x \cos ^{2} x$$=\cos ^{2} 2 x+\frac{1}{2} \sin ^{2} 2 x$原式 $=\int \frac{\mathrm{d} x}{\cos ^{2} 2 x+\frac{1}{2} \sin ^{2} 2 x}$
设f(x)是可导函数,且f′(x)=sin<sup>2</sup>[sin(x+1)],f(0)=4,f(x)的反函数是x=φ(y),则φ′(4)=()。 A: 1/sin<sup>2</sup>(sin1) B: sin<sup>2</sup>(sin1) C: -sin<sup>2</sup>(sin1) D: -1/sin<sup>2</sup>(sin1)
设f(x)是可导函数,且f′(x)=sin<sup>2</sup>[sin(x+1)],f(0)=4,f(x)的反函数是x=φ(y),则φ′(4)=()。 A: 1/sin<sup>2</sup>(sin1) B: sin<sup>2</sup>(sin1) C: -sin<sup>2</sup>(sin1) D: -1/sin<sup>2</sup>(sin1)
y=arcsin(4x+1)的反函数为 A: y=(sinx-1)/4, x∈R B: y=sin[(x-1)/4], x∈R C: y=sin[(x-1)/4], x∈[-π/2,π/2] D: y=(sinx-1)/4, x∈[-π/2,π/2]
y=arcsin(4x+1)的反函数为 A: y=(sinx-1)/4, x∈R B: y=sin[(x-1)/4], x∈R C: y=sin[(x-1)/4], x∈[-π/2,π/2] D: y=(sinx-1)/4, x∈[-π/2,π/2]
智慧职教: 某一端口网络的电压和电流为关联参考方向,其电压和电流分别为: u(t)=16+25√2sinωt+4√2 sin(3ωt+30°)+√6 sin(5ωt+50°)V ,i(t)=3+10√2 sin(ωt-60°)+4√2 sin(2ωt+20°)+2√2 sin(4ωt+40°) A,则网络的平均功率_______。
智慧职教: 某一端口网络的电压和电流为关联参考方向,其电压和电流分别为: u(t)=16+25√2sinωt+4√2 sin(3ωt+30°)+√6 sin(5ωt+50°)V ,i(t)=3+10√2 sin(ωt-60°)+4√2 sin(2ωt+20°)+2√2 sin(4ωt+40°) A,则网络的平均功率_______。
1、已知cos4α=2/3,则(sin^4α-cos^4α)^2=2、(1-cosx+sinx)/(1+cosx+sinx)=-2,则tanx=
1、已知cos4α=2/3,则(sin^4α-cos^4α)^2=2、(1-cosx+sinx)/(1+cosx+sinx)=-2,则tanx=