Current location - Training Enrollment Network - Mathematics courses - Mathematical problems in higher mathematics
Mathematical problems in higher mathematics
Second pass: make y/x = t = = = > y = XT = = =>;; dy/dx = t + x dt/dx

t + x dt/dx = t? Equivalent to -ED

x dt/dx = t? - 2t

dt / (t? -2t) = dx /x usage 1/ (t? -2t)= 1/2[ 1/(t-2)- 1/t]

dt/[ 1/(t-2)- 1/t]= 2dx/x

Ln((t-2) / t) = Lnx? + LnC

(t-2) / t = Cx?

t = -2 / (Cx? - 1)

That is, y/x = -2/(Cx? - 1)

y = -2x / (Cx? - 1)

When the substitution condition is x = 1, y( 1) = 1, C =-1 is obtained.

The special solution is: y = 2x/(x? + 1)

The third way: find the double integral ∫∫√ x 2+y 2dxddy, where the integral area d = {(x, y) | x 2+y 2 ≤ 2x, 0≤y≤x}.

Reference answer: 10/9√2

D={(x,y)|x? +y? ≤2x,0≤y≤x}

= = = & gtD={(x,y)|(x- 1)? + y? ≤ 1,0≤y≤x}

= = => method 1: x before Y: D = {(X, Y) | (X = Y → 1+√ (1-Y? ),0≤y≤ 1}

Method 2: polar coordinates: r? = 2COS θ, that is, θ = 0→π/4; r = 0→2cosθ

Original integral = ∫∫√ x 2+y 2dxddy

=∫{θ = 0→π/4}∫{r = 0→2cosθ} √r? r drdθ

=∫{θ = 0→π/4}∫{r = 0→2cosθ} r? drdθ

=∫{θ = 0→π/4} 8cos? θ /3 dθ

=8/3∫{θ = 0→π/4} 1 - sin? θ dsinθ

=8/3 [ sinθ - sin? θ/3 ] {sinθ = 0→ 1/√2}

=8/3 [ 1/√2 - 1/(6√2)]

=8/3 [5/(6√2)]

= 10/(9√2)