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1 Given the initial condition y(1) = 100, what is the particular solution of the equation 1. = 34232 l 3) 2 3- O y
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Given the initial condition y(1) = 100, what is the particular solution of the equation 1." = 34232 l 3) 2 3- O y = e5 (32\") 2 0.483 2 O y = 0.48333 (32\") 3 NIH O y = 0.48333 (9+3) 3 O y = 6% (32\") 2 l 0.483 Rewrite the following equation with the variables separated. 3vt+ v2 dy dt = 6 O y? dy = (6 +3vt) dt O y? dy = (6 -3vt) dt O y? dy - 6 = (3vt) dt O y? dy - 3 = (3vt ) dtRewrite the following equation with the variables separated. 3xy' + 2y = sin(y). dy O (sin( y )-2y) 3 1 dy O = 3x dx (sin (y )-2y) O (sin(y) - 2y) dy = 3x dx dy = O (sin( y ) + 2y) 3 1Rewrite the following equation with the variables separated. = 4 12(e# +2 ) 1 O dy = 4x'dx ( ev + 2 ) O (ey + 2 ) dy = 1 dx 4x2 1 O (ev+2) y = 4x2 dx O (ey +2) dy = 4x'dxWhat is the general solution to (x2 + 2) dy = 2yx? O y = In (a2 + 2) + C oy = -2 +C O y= kx2+ C O y = x2 + CWhat is the general solution to y y' + sin(x) - 3 = 0? O y= V9x + 3 cos(x) + C o y= 35x + cos(x) + C oy= 9x + 3 cos(x) + C O y= 19x -3 cos(x) + CWhat is the general solution to (a2 + 1) dy =x(y+1) O y= evx-+1 - C O y = Cvx2+1-1 O y = Cer +1 _ tok O y = CVx2+1+192 What is the equation of the curve that passes through the point (1, 4)and has a slope of at any point (2:, y)? 1/:+3 4.2521/33 O _ 1 y _ 4.252'/z+3 O _ 1 y _ 3.2521/z+3 o = 1 y 3.25+\\/2z+3 dy Given the initial condition y(0) = 2, what is the particular solution of the equation va - vy = 0? o y =1 + 1.587 Oy= 23 +2.828 Oy= 22 - 1.587 Oy= 12+2.828Step by Step Solution
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