Consider the fifferential equation dy/dx=(3-y)cosx.Let y=f(x)be the particular solution to the differential equation with the inital condition f(0)=1.
dy / dx = ( 3 - y ) cos x
Divide both sides by 3 - y
dy / [ dx ∙ ( 3 - y ) ] = cos x
Multiply both sides by dx
dy / ( 3 - y ) = cos x dx
Substitution:
3 - y = u
- dy = du
dy = - du
∫ dy / ( 3 - y ) = ∫ - du / u = - ∫ du / u = - log ( u ) + C₁ = - log ( 3 - y ) + C₁
∫ cos x dx = sin x + C₂
∫ dy / ( 3 - y ) = ∫ cos x dx
- log ( 3 - y ) + C₁ = sin x + C₂
Subtract C₁ to both sides
- log ( 3 - y ) = sin x + C₂ - C₁
C₂ - C₁ = C
- log ( 3 - y ) = sin x + C
Multiply both sides by - 1
log ( 3 - y ) = - sin x - C
e^[ log ( 3 - y ) ] = e^( - sin x - C )
3 - y = e^(- sin x ) ∙ e^( - C )
Subtract 3 to both sides
- y = e^( - sin x ) ∙ e^( - C ) - 3
Multiply both sides by - 1
y = - e^( - sin x ) ∙ e^( - C ) + 3
x = 0 , y = 1
y = - e^( - sin x ) ∙ e^( - C ) + 3
1 = - e^( - sin 0 ) ∙ e^( - C ) + 3
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- sin 0 = - ( 0 ) = 0
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1 = - e^0 ∙ e^( - C ) + 3
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e^0 = 1
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1 = - 1 ∙ e^( - C ) + 3
1 = - e^( - C ) + 3
Add e^( - C ) to both sides
1 + e^( - C ) = - e^( - C ) + e^( - C ) + 3
1 + e^( - C ) = 3
Subtract 1 to both sides
e^( - C ) = 2
y = - e^( - sin x ) ∙ e^( - C ) + 3
y = - e^( - sin x ) ∙ 2 + 3
y = - 2 e^( - sin x ) + 3
To solve the given differential equation, we can separate the variables:
dy/(3-y) = cos(x) dx
Now, we can integrate both sides of the equation:
∫(1/(3-y)) dy = ∫cos(x) dx
To find the integral of 1/(3-y), we can use the substitution method. Let u = 3-y, therefore du = -dy. Substituting these values, the equation becomes:
-∫(1/u) du = ∫cos(x) dx
Simplifying further:
-ln|u| = ∫cos(x) dx
Now, we can find the integral of cos(x), which is sin(x):
-ln|u| = sin(x) + C_1
where C_1 is the constant of integration.
Substituting back the value of u = 3-y:
-ln|3-y| = sin(x) + C_1
To solve for y, we need to isolate y. We can start by getting rid of the absolute value:
ln|3-y| = -sin(x) - C_1
Taking the exponential of both sides:
|3-y| = e^(-sin(x) - C_1)
Now, we can remove the absolute value sign by considering two cases:
1. For 3-y > 0:
3-y = e^(-sin(x) - C_1)
2. For 3-y < 0:
y-3 = e^(-sin(x) - C_1)
Now, let's solve these cases separately:
1. For 3-y > 0, we have:
3-y = e^(-sin(x) - C_1)
Simplifying further:
y = 3 - e^(-sin(x) - C_1)
2. For 3-y < 0, we have:
y-3 = e^(-sin(x) - C_1)
Simplifying further:
y = 3 + e^(-sin(x) - C_1)
Finally, we can apply the initial condition f(0) = 1 to determine the value of the constant C_1.
When x = 0, y = f(0) = 1.
Substituting these values into the general solution:
1 = 3 - e^(-sin(0) - C_1)
Simplifying:
1 = 3 - e^(0 - C_1)
1 = 3 - e^(-C_1)
e^(-C_1) = 2
Taking the natural logarithm of both sides:
-C_1 = ln(2)
C_1 = -ln(2)
Therefore, the particular solution to the given differential equation with the initial condition f(0) = 1 is:
y = 3 - e^(-sin(x) + ln(2))
y = 3 - (2 * e^(-sin(x)))
To find the particular solution to the given differential equation with the initial condition, we can follow these steps:
Step 1: Separate the variables.
dy/(3 - y) = cos(x) dx
Step 2: Integrate both sides of the equation.
∫(1/(3 - y)) dy = ∫cos(x) dx
Step 3: Evaluate the integrals.
ln|3 - y| = sin(x) + C
Step 4: Solve for y.
|3 - y| = e^(sin(x) + C)
Step 5: Eliminate the absolute value.
Case 1: 3 - y > 0
3 - y = e^(sin(x) + C)
y = 3 - e^(sin(x) + C)
Case 2: 3 - y < 0
y - 3 = e^(sin(x) + C)
y = 3 + e^(sin(x) + C)
Step 6: Use the initial condition to find the constant C.
Given f(0) = 1, substitute x = 0, y = 1 into the equation.
1 = 3 - e^(sin(0) + C)
2 = e^(C)
C = ln(2)
Step 7: Substitute the value of C back into the equation.
For Case 1: y = 3 - e^(sin(x) + ln(2))
For Case 2: y = 3 + e^(sin(x) + ln(2))
Therefore, the particular solution to the differential equation with the initial condition f(0) = 1 is:
y = 3 - e^(sin(x) + ln(2)) or y = 3 + e^(sin(x) + ln(2))