Solve Q8 plece Q1) One of the following is correct for functions the f(x) = 1, g(x) = x³ and h(x) = ln(x):a) f(x), g(x) and h(x) are linearly dependentc) W (f(x), g(x), h(x)) = 0b) f(x), g(x) and h(x) are linearly independentd) W (f(x), g(x), h(x)) = 9Q2) L-1-2sS²+7²)=a) -2coshπtb) -2sinhntc)-2сostd) -2sinntQ3) The Integrating factor which make (3x²y + 2xy + y³) dx + (x² + y²)dy = 0 exact, is:a) e-3xb) exc)exd) e 3xQ4) The linear form of nonlinear ODE y' - 2y = 2y, is:a) u' + 6u = -6b) u' - 6u = -6c) u' - 6u = 6d) u' + 6u = 6Q5) The general solution of 2x2y" + 3xy' - 15y = 0, is:a) y(x) = C₁x² + ₂x³b) y(x) = ₂x + ₂x-³5c) y(x) = C₁x² + ₂x−3d) y(x) = C₁xz+C₂x³Q6) Evaluate(2 e-2t sin4t -0.5 cos3t):8Sb)8(S-2)²+16 2s²+184(s+2)² +162s²+18c) (8+2)²+16 28² +18(S-2)²+16 2s²+18Q7) The general solution of y"-4y' +9y = 0, is:a) y(t) = c₁e²t cos(5 t) + c₂e²t sin (5 t)c)y(t)=c₁e²t cos(√5 t) + c₂e²tsin (√5 t)b)y(t)=c₁et cos(√5 t) + c₂etsin (√5t)d) y(t) = c₁et cos(5 t) + c₂esin (5 t)Q8) The inverse Laplacetransform of H(s):=is:I(3s+2)(S-2)e2t|a) ƒ (t) = ¹ + ²t| c) ƒ (t) ==³e² +²e² -2t181b) f(t) ==e+e-²t+²e²tеd)f(t) = ²³²e:BQ9) The solution of y" + y' = 0 by using power series method, is:x3 x²a) y(x) = a₁ + a₁( 1 −²+2.-...)b) y(x) = ao + a₁(1+²415!x5c) y(x) = a₁ + α₁ ( x −₁+3+.)d)y(x) = ao+a₁(x+²4! 513!IT·+...)- ...)

Question

Solve Q8 plece Q1) One of the following is correct for functions the f(x) = 1, g(x) = x³ and h(x) = ln(x):a) f(x), g(x) and h(x) are linearly dependentc) W (f(x), g(x), h(x)) = 0b) f(x), g(x) and h(x) are linearly independentd) W (f(x), g(x), h(x)) = 9Q2) L-1-2sS²+7²)=a) -2coshπtb) -2sinhntc)-2сostd) -2sinntQ3) The Integrating factor which make (3x²y + 2xy + y³) dx + (x² + y²)dy = 0 exact, is:a) e-3xb) exc)exd) e 3xQ4) The linear form of nonlinear ODE y' - 2y = 2y, is:a) u' + 6u = -6b) u' - 6u = -6c) u' - 6u = 6d) u' + 6u = 6Q5) The general solution of 2x2y&#34; + 3xy' - 15y = 0, is:a) y(x) = C₁x² + ₂x³b) y(x) = ₂x + ₂x-³5c) y(x) = C₁x² + ₂x−3d) y(x) = C₁xz+C₂x³Q6) Evaluate(2 e-2t sin4t -0.5 cos3t):8Sb)8(S-2)²+16 2s²+184(s+2)² +162s²+18c) (8+2)²+16 28² +18(S-2)²+16 2s²+18Q7) The general solution of y&#34;-4y' +9y = 0, is:a) y(t) = c₁e²t cos(5 t) + c₂e²t sin (5 t)c)y(t)=c₁e²t cos(√5 t) + c₂e²tsin (√5 t)b)y(t)=c₁et cos(√5 t) + c₂etsin (√5t)d) y(t) = c₁et cos(5 t) + c₂esin (5 t)Q8) The inverse Laplacetransform of H(s):=is:I(3s+2)(S-2)e2t|a) ƒ (t) = ¹ + ²t| c) ƒ (t) ==³e² +²e² -2t181b) f(t) ==e+e-²t+²e²tеd)f(t) = ²³²e:BQ9) The solution of y&#34; + y' = 0 by using power series method, is:x3 x²a) y(x) = a₁ + a₁( 1 −²+2.-...)b) y(x) = ao + a₁(1+²415!x5c) y(x) = a₁ + α₁ ( x −₁+3+.)d)y(x) = ao+a₁(x+²4! 513!IT·+...)- ...)

Accepted Answer

Answered: Image /qna-images/answer/a3ecc126-b15f-46ef-bbcc-bf9be65d5701.jpg

The inverse Laplace transform of H(s) = + (t) ==²e²+e-²t 8 1 " (3s+2)(S-2) b) f(t) ==e+e-2t -3 d)f (t) == +1 +²e²t is: