Question

find the general solution to y'''-y''+y'-y=0​

Answer 1

Answer:

Step 1:

Assume a solution will be proportional to e^(λ x) for some constant λ.

Step 2:

Substitute y(x) = e^(λ x) into the differential equation:

( d^2 )/( dx^2)(e^(λ x)) + e^(λ x) = 0

Substitute ( d^2 )/( dx^2)(e^(λ x)) = λ^2 e^(λ x):

λ^2 e^(λ x) + e^(λ x) = 0

Step 3:

Factor out e^(λ x):

(λ^2 + 1) e^(λ x) = 0

Step 4:

Since e^(λ x) !=0 for any finite λ, the zeros must come from the polynomial:

λ^2 + 1 = 0

Step 5:

Solve for λ:

λ = i or λ = -i

Step 6:

The roots λ = ± i give y_1(x) = c_1 e^(i x), y_2(x) = c_2 e^(-i x) as solutions, where c_1 and c_2 are arbitrary constants.

The general solution is the sum of the above solutions:

y(x) = y_1(x) + y_2(x) = c_1 e^(i x) + c_2 e^(-i x)

Step 7:

Apply Euler's identity e^(α + i β) = e^α cos(β) + i e^α sin(β):

y(x) = c_1 (cos(x) + i sin(x)) + c_2 (cos(x) - i sin(x))

Step 8:

Regroup terms:

y(x) = (c_1 + c_2) cos(x) + i (c_1 - c_2) sin(x)

Redefine c_1 + c_2 as c_1 and i (c_1 - c_2) as c_2, since these are arbitrary constants:

Answer: y(x) = c_1 cos(x) + c_2 sin(x)

Answer 2

The General equation is y = k1$e^{x} +k2cosx+k3sinx$

Given:

We have the differential equation y'''-y''+y'-y=0​

To Find:

To get the value of the  arbitrary constants using the general solution

Solution:

We have the differential equation,

y'''-y''+y'-y=0​

We immediately see that it is second-order homogeneous linear DE with constant coefficients. The usual technique for this type of D.E is to first suppose that there is a solution of the form

y = $e^{mx}$

Differentiate this supposition with respect to x, then we have,

$y^{'} = me^{mx}$ and,

$y^{''} = m^2e^{mx}$ and,

$y^{'''} = m^3e^{mx}$

Substitute into the original differential equation then we obtain,

$m^3e^{mx} -$ $m^2e^{mx}$ +m $e^{mx}$ - $e^{mx}$

$e^{mx}$ ( $m^{3} - m^{2} +m -1$) =0

Since , $e^{mx}$ can not be equal 0, then we have,

$m^{3} - m^{2} +m -1$ =0

Then, (m-1)(m^2+1) = 0

Then the roots are,

m1 =1

$m_{2,3} = +i,-i$

m2 = +i ,m3 = -i

which are one real, two conjugate complex and distinct roots.

Then using theorem and theorem, the general solution of the differential equation,

$y^{''} +6y^{'}+13y = 0$

where k,c1,c2  are an arbitrary constants

Which simply is,

y = k1$e^{x} +k2cosx+k3sinx$

Hence, we get the arbitrary constants as  y = k1$e^{x} +k2cosx+k3sinx$

#SPJ2

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