Recall that variation of parameters is used to solve second-order ODEs of the form
<em>y''(t)</em> + <em>p(t)</em> <em>y'(t)</em> + <em>q(t)</em> <em>y(t)</em> = <em>f(t)</em>
so the first thing you need to do is divide both sides of your equation by <em>t</em> :
<em>y''</em> + (2<em>t</em> - 1)/<em>t</em> <em>y'</em> - 2/<em>t</em> <em>y</em> = 7<em>t</em>
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You're looking for a solution of the form

where


and <em>W</em> denotes the Wronskian determinant.
Compute the Wronskian:

Then


The general solution to the ODE is

which simplifies somewhat to

Answer:
x = 17
Step-by-step explanation:
Since the triangles are similar then corresponding angles are congruent.
∠ I = ∠P ← substitute values and solve for x
3x + 4 = 72 - x ( add x to both sides )
4x + 4 = 72 ( subtract 4 from both sides )
4x = 68 ( divide both sides by 4 )
x = 17
Answer:
145
Step-by-step explanation:
Divide 48 by 7 = 6≅. then add 6 to 139. 145
The answer fam is....... 3x - 10