Q7 Q8 What is the domain of the solution of the first order linear initial value
ID: 1890064 • Letter: Q
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Q7
Q8
What is the domain of the solution of the first order linear initial value problem: dy / dx + 3 / 5x - 8 y = - 9 / 4x - 7 with y = - 4 when x= -2. The domain of the solution is ( , ). Is the matrix invertible? Your answer (yes or no) After multiplying through by the integrating factor, the linear differential equation, dy / dx + 6 / 6x + 5 y = tan (-5x), with initial condition y = 3 when x = 1, has the modified form d / dx ((6x + 5)?y) = (6x + 5) middot tan(-5x). Give the exact answer to the appropriate definite integral of the Left Hand Side of the modified differential equation. The answer to the appropriate integral of the left hand side is .Explanation / Answer
q6) dy/dx + 3y/(5x-8) = -9/(4x-7) Solve the linear equation ( dy(x))/( dx)+(3 y(x))/(5 x-8) = -9/(4 x-7): Let mu(x) = exp( integral 3/(5 x-8) dx) = (5 x-8)^(3/5). Multiply both sides by mu(x): (5 x-8)^(3/5) ( dy(x))/( dx)+(3 y(x))/(5 x-8)^(2/5) = -(9 (5 x-8)^(3/5))/(4 x-7) Substitute 3/(5 x-8)^(2/5) = ( d)/( dx)((5 x-8)^(3/5)): (5 x-8)^(3/5) ( dy(x))/( dx)+( d)/( dx)((5 x-8)^(3/5)) y(x) = -(9 (5 x-8)^(3/5))/(4 x-7) Apply the reverse product rule f ( dg)/( dx)+( df)/( dx) g = ( d)/( dx)(f g) to the left-hand side: ( d)/( dx)((5 x-8)^(3/5) y(x)) = -(9 (5 x-8)^(3/5))/(4 x-7) Integrate both sides with respect to x: integral ( d)/( dx)((5 x-8)^(3/5) y(x)) dx = integral -(9 (5 x-8)^(3/5))/(4 x-7) dx Evaluate the integrals: (5 x-8)^(3/5) y(x) = 3/64 (-12 2^(4/5) 3^(3/5) log(-2^(2/5) 3^(4/5) (5 x-8)^(1/5)+3)-3 2^(4/5) 3^(3/5) (-1+sqrt(5)) log(1/2 (2 2^(4/5) 3^(3/5) (5 x-8)^(2/5)+2^(2/5) 3^(4/5) (1+sqrt(5)) (5 x-8)^(1/5)+6))+3 2^(4/5) 3^(3/5) (1+sqrt(5)) log(1/2 (2 2^(4/5) 3^(3/5) (5 x-8)^(2/5)-2^(2/5) 3^(4/5) (-1+sqrt(5)) (5 x-8)^(1/5)+6))-12 2^(3/10) 3^(3/5) sqrt(5-sqrt(5)) tan^(-1)((4 2^(2/5) 3^(4/5) (5 x-8)^(1/5)+3-3 sqrt(5))/(3 sqrt(2) sqrt(5+sqrt(5))))+12 2^(3/10) 3^(3/5) sqrt(5+sqrt(5)) tan^(-1)((4 2^(2/5) 3^(4/5) (5 x-8)^(1/5)+3+3 sqrt(5))/(3 sqrt(10-2 sqrt(5))))-80 (5 x-8)^(3/5))+c_1, where c_1 is an arbitrary constant. Divide both sides by mu(x) = (5 x-8)^(3/5): y(x) = (3/64 (-12 2^(4/5) 3^(3/5) log(-2^(2/5) 3^(4/5) (5 x-8)^(1/5)+3)-3 2^(4/5) 3^(3/5) (-1+sqrt(5)) log(1/2 (2 2^(4/5) 3^(3/5) (5 x-8)^(2/5)+2^(2/5) 3^(4/5) (1+sqrt(5)) (5 x-8)^(1/5)+6))+3 2^(4/5) 3^(3/5) (1+sqrt(5)) log(1/2 (2 2^(4/5) 3^(3/5) (5 x-8)^(2/5)-2^(2/5) 3^(4/5) (-1+sqrt(5)) (5 x-8)^(1/5)+6))-12 2^(3/10) 3^(3/5) sqrt(5-sqrt(5)) tan^(-1)((4 2^(2/5) 3^(4/5) (5 x-8)^(1/5)+3-3 sqrt(5))/(3 sqrt(2) sqrt(5+sqrt(5))))+12 2^(3/10) 3^(3/5) sqrt(5+sqrt(5)) tan^(-1)((4 2^(2/5) 3^(4/5) (5 x-8)^(1/5)+3+3 sqrt(5))/(3 sqrt(10-2 sqrt(5))))-80 (5 x-8)^(3/5))+c_1)/(5 x-8)^(3/5) check the values of x possible. It is the Domain q7) for given Matrix A |A| is not equal to zero. so it is invertible q8) d[(6x+5)y]/dx = (6x+5)* tan(-5x) By integration we get 6y = -(6x+5) * tan(5x)
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