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What is the equation of a curve passing through (0, 1) and whose differential equation is given by dy = y tan x dx ?
- a)y = cosx
- b)y = sinx
- c)y = secx
- d)y = cosec x
Correct answer is option 'C'. Can you explain this answer?
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To find the equation of the curve passing through (0, 1) with the given differential equation, we need to integrate the differential equation and solve for y.
Given: dy = y tan(x) dx
Integration:
Integrating both sides with respect to x, we get:
∫ dy = ∫ y tan(x) dx
Simplifying the integral on the left side gives us:
y = ∫ y tan(x) dx
Now, let's solve the integral on the right side. We can use integration by parts to evaluate the integral. The formula for integration by parts is:
∫ u dv = uv - ∫ v du
In this case, we can choose u = y and dv = tan(x) dx. Taking the derivatives and integrals, we have:
du = dy
v = ∫ tan(x) dx = -ln|cos(x)|
Applying the formula for integration by parts, we get:
∫ y tan(x) dx = y (-ln|cos(x)|) - ∫ -ln|cos(x)| dy
Simplifying further, we have:
∫ y tan(x) dx = -y ln|cos(x)| + ∫ ln|cos(x)| dy
Now, we can substitute this back into the original equation:
y = -y ln|cos(x)| + ∫ ln|cos(x)| dy
Solving the Integral:
To solve the integral on the right side, we integrate with respect to y:
y = -y ln|cos(x)| + y ln|y| + C
where C is the constant of integration.
Using Initial Condition:
To find the value of C, we can use the initial condition given in the problem, which states that the curve passes through (0, 1). Substituting these values into the equation, we have:
1 = -1 ln|cos(0)| + 1 ln|1| + C
1 = -1 ln|1| + C
1 = C
Substituting C = 1 back into the equation, we get the final equation of the curve:
y = -y ln|cos(x)| + y ln|y| + 1
Simplifying further, we have:
y + y ln|cos(x)| - y ln|y| = 1
Given: dy = y tan(x) dx
Integration:
Integrating both sides with respect to x, we get:
∫ dy = ∫ y tan(x) dx
Simplifying the integral on the left side gives us:
y = ∫ y tan(x) dx
Now, let's solve the integral on the right side. We can use integration by parts to evaluate the integral. The formula for integration by parts is:
∫ u dv = uv - ∫ v du
In this case, we can choose u = y and dv = tan(x) dx. Taking the derivatives and integrals, we have:
du = dy
v = ∫ tan(x) dx = -ln|cos(x)|
Applying the formula for integration by parts, we get:
∫ y tan(x) dx = y (-ln|cos(x)|) - ∫ -ln|cos(x)| dy
Simplifying further, we have:
∫ y tan(x) dx = -y ln|cos(x)| + ∫ ln|cos(x)| dy
Now, we can substitute this back into the original equation:
y = -y ln|cos(x)| + ∫ ln|cos(x)| dy
Solving the Integral:
To solve the integral on the right side, we integrate with respect to y:
y = -y ln|cos(x)| + y ln|y| + C
where C is the constant of integration.
Using Initial Condition:
To find the value of C, we can use the initial condition given in the problem, which states that the curve passes through (0, 1). Substituting these values into the equation, we have:
1 = -1 ln|cos(0)| + 1 ln|1| + C
1 = -1 ln|1| + C
1 = C
Substituting C = 1 back into the equation, we get the final equation of the curve:
y = -y ln|cos(x)| + y ln|y| + 1
Simplifying further, we have:
y + y ln|cos(x)| - y ln|y| = 1
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What is the equation of a curve passing through (0, 1) and whose differential equation is given by dy = y tan x dx ?a)y = cosxb)y = sinxc)y = secxd)y = cosec xCorrect answer is option 'C'. Can you explain this answer?
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