The first step for deriving the quadratic formula from the quadratic equation, 0 = ax2 + bx + c, is shown. Step 1: –c = ax2 + bx
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Answer:
The solution
Step-by-step explanation:
<u><em>Explanation</em></u>:-
Consider the initial value problem y′+3 y=9 t,y(0)=7
<em>Step(i)</em>:-
Given differential problem
y′+3 y=9 t
<em>Take the Laplace transform of both sides of the differential equation</em>
L( y′+3 y) = L(9 t)
<em>Using Formula Transform of derivatives</em>
<em> L(y¹(t)) = s y⁻(s)-y(0)</em>
<em> By using Laplace transform formula</em>
<em> </em><em> </em>
<em>Step(ii):-</em>
Given
L( y′(t)) + 3 L (y(t)) = 9 L( t)
Taking common y⁻(s) and simplification, we get
<em>Step(iii</em>):-
<em>By using partial fractions , we get</em>
On simplification we get
9 = A s(s+3) +B(s+3) +C(s²) ...(i)
Put s =0 in equation(i)
9 = B(0+3)
<em> B = 9/3 = 3</em>
Put s = -3 in equation(i)
9 = C(-3)²
<em>C = 1</em>
Given Equation 9 = A s(s+3) +B(s+3) +C(s²) ...(i)
Comparing 'S²' coefficient on both sides, we get
9 = A s²+3 A s +B(s)+3 B +C(s²)
<em> 0 = A + C</em>
<em>put C=1 , becomes A = -1</em>
<u><em>Step(iv):-</em></u>
Applying inverse Laplace transform on both sides
<em>By using inverse Laplace transform</em>
<em></em><em></em>
<u><em>Final answer</em></u>:-
<em>Now the solution , we get</em>