Answer :
Sure, let's solve the problem step by step.
1. Given Function and Its Derivative:
We start with the given function [tex]\( y = a + b t + c t^2 \)[/tex].
2. Find the Derivative:
Calculate the first derivative of [tex]\( y \)[/tex] with respect to [tex]\( t \)[/tex]:
[tex]\[ y' = \frac{d}{dt}(a + b t + c t^2) = b + 2c t. \][/tex]
3. Substitute [tex]\( y \)[/tex] and [tex]\( y' \)[/tex] into the given equation:
The given equation is [tex]\( y' + y = 1 + t^2 \)[/tex]. Substituting [tex]\( y \)[/tex] and [tex]\( y' \)[/tex] we get:
[tex]\[ (b + 2c t) + (a + b t + c t^2) = 1 + t^2. \][/tex]
4. Combine Like Terms:
Combine the terms involving [tex]\( t \)[/tex]:
[tex]\[ a + b + b t + 2c t + c t^2 = 1 + t^2. \][/tex]
Simplify the equation:
[tex]\[ a + b + (b + 2c)t + c t^2 = 1 + t^2. \][/tex]
5. Equate Coefficients:
For the equation to hold for all [tex]\( t \)[/tex], the coefficients of the corresponding powers of [tex]\( t \)[/tex] must be equal. Hence, we set up the following system of equations by matching the coefficients:
Comparing the constant terms:
[tex]\[ a + b = 1. \][/tex]
Comparing the coefficients of [tex]\( t \)[/tex]:
[tex]\[ b + 2c = 0. \][/tex]
Comparing the coefficients of [tex]\( t^2 \)[/tex]:
[tex]\[ c = 1. \][/tex]
6. Solve the System of Equations:
From the third equation, we already know:
[tex]\[ c = 1. \][/tex]
Substitute [tex]\( c = 1 \)[/tex] into the second equation:
[tex]\[ b + 2(1) = 0 \implies b + 2 = 0 \implies b = -2. \][/tex]
Now substitute [tex]\( b = -2 \)[/tex] into the first equation:
[tex]\[ a + (-2) = 1 \implies a - 2 = 1 \implies a = 3. \][/tex]
7. Find Particular Solution:
Therefore, the values of [tex]\( a, b, \)[/tex] and [tex]\( c \)[/tex] that satisfy the given differential equation are:
[tex]\[ a = 3, \quad b = -2, \quad c = 1. \][/tex]
The particular solution is found by substituting these values back into the general form of [tex]\( y \)[/tex]:
[tex]\[ y = 3 - 2t + t^2. \][/tex]
Thus, the particular solution to [tex]\( y' + y = 1 + t^2 \)[/tex] with [tex]\( y = a + b t + c t^2 \)[/tex] is:
[tex]\[ y = 3 - 2 t + t^2. \][/tex]
1. Given Function and Its Derivative:
We start with the given function [tex]\( y = a + b t + c t^2 \)[/tex].
2. Find the Derivative:
Calculate the first derivative of [tex]\( y \)[/tex] with respect to [tex]\( t \)[/tex]:
[tex]\[ y' = \frac{d}{dt}(a + b t + c t^2) = b + 2c t. \][/tex]
3. Substitute [tex]\( y \)[/tex] and [tex]\( y' \)[/tex] into the given equation:
The given equation is [tex]\( y' + y = 1 + t^2 \)[/tex]. Substituting [tex]\( y \)[/tex] and [tex]\( y' \)[/tex] we get:
[tex]\[ (b + 2c t) + (a + b t + c t^2) = 1 + t^2. \][/tex]
4. Combine Like Terms:
Combine the terms involving [tex]\( t \)[/tex]:
[tex]\[ a + b + b t + 2c t + c t^2 = 1 + t^2. \][/tex]
Simplify the equation:
[tex]\[ a + b + (b + 2c)t + c t^2 = 1 + t^2. \][/tex]
5. Equate Coefficients:
For the equation to hold for all [tex]\( t \)[/tex], the coefficients of the corresponding powers of [tex]\( t \)[/tex] must be equal. Hence, we set up the following system of equations by matching the coefficients:
Comparing the constant terms:
[tex]\[ a + b = 1. \][/tex]
Comparing the coefficients of [tex]\( t \)[/tex]:
[tex]\[ b + 2c = 0. \][/tex]
Comparing the coefficients of [tex]\( t^2 \)[/tex]:
[tex]\[ c = 1. \][/tex]
6. Solve the System of Equations:
From the third equation, we already know:
[tex]\[ c = 1. \][/tex]
Substitute [tex]\( c = 1 \)[/tex] into the second equation:
[tex]\[ b + 2(1) = 0 \implies b + 2 = 0 \implies b = -2. \][/tex]
Now substitute [tex]\( b = -2 \)[/tex] into the first equation:
[tex]\[ a + (-2) = 1 \implies a - 2 = 1 \implies a = 3. \][/tex]
7. Find Particular Solution:
Therefore, the values of [tex]\( a, b, \)[/tex] and [tex]\( c \)[/tex] that satisfy the given differential equation are:
[tex]\[ a = 3, \quad b = -2, \quad c = 1. \][/tex]
The particular solution is found by substituting these values back into the general form of [tex]\( y \)[/tex]:
[tex]\[ y = 3 - 2t + t^2. \][/tex]
Thus, the particular solution to [tex]\( y' + y = 1 + t^2 \)[/tex] with [tex]\( y = a + b t + c t^2 \)[/tex] is:
[tex]\[ y = 3 - 2 t + t^2. \][/tex]