When sedimentary rock is exposed to heat and pressure it changes into METAMORPHIC ROCK.
Heat and pressure have the capacity to change a rock into a completely new type of rock. An igneous rock or a sedimentary rock can be changed into a metamorphic rock as a result of heat and pressure which the rock is subjected to. Metamorphic rocks are usually formed from already existing rocks that are exposed to pressure and heat.
Answer:
The heating of saturated solutions lead to the increase in their solubility. The kinetic energy of the liquid molecules increases on increasing the temperature due to which they move apart from each other creating more space for the solute molecules to dissolve thereby increasing the solubility.
Yp(t) = A1 t^2 + A0 t + B0 t e(4t)
=> y ' = 2A1t + A0 + B0 [e^(4t) +4 te^(4t) ]
y ' = 2A1t + A0 + B0e^(4t) + 4B0 te^(4t)
=> y '' = 2A1 + 4B0e(4t) + 4B0 [ e^(4t) + 4te^(4t)
y '' = 2A1 + 4B0e^(4t) + 4B0e^(4t) + 16B0te^(4t)
Now substitute the values of y ' and y '' in the differential equation:
<span>y′′+αy′+βy=t+e^(4t)
</span> 2A1 + 4B0e^(4t) + 4B0e^(4t) + 16B0te^(4t) + α{2A1t + A0 + B0e^(4t) + 4B0 te^(4t) } + β{A1 t^2 + A0 t + B0 t e(4t)} = t + e^(4t)
Next, we equate coefficients
1) Constant terms of the left side = constant terms of the right side:
2A1+ 2αA0 = 0 ..... eq (1)
2) Coefficients of e^(4t) on both sides
8B0 + αB0 = 1 => B0 (8 + α) = 1 .... eq (2)
3) Coefficients on t
2αA1 + βA0 = 1 .... eq (3)
4) Coefficients on t^2
βA1 = 0 ....eq (4)
given that A1 ≠ 0 => β =0
5) terms on te^(4t)
16B0 + 4αB0 + βB0 = 0 => B0 (16 + 4α + β) = 0 ... eq (5)
Given that B0 ≠ 0 => 16 + 4α + β = 0
Use the value of β = 0 found previously
16 + 4α = 0 => α = - 16 / 4 = - 4.
Answer: α = - 4 and β = 0
