Question

Let y = y₁(x) and y = y₂(x) be two distinct solution of the differential equation
\(\frac{dy}{dx} = x+y,\)
with y₁(0) = 0 and y₂(0) = 1 respectively. Then, the number of points of intersection of y = y₁ (x) and y = y₂(x) is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The Correct Option is A

To determine the number of points of intersection of the two solutions \( y = y_1(x) \) and \( y = y_2(x) \) of the differential equation \( \frac{dy}{dx} = x + y \), we proceed as follows:

1. First, solve the given first-order linear differential equation \( \frac{dy}{dx} = x + y \). This equation can be rearranged as \( \frac{dy}{dx} - y = x \). 
2. This is a linear differential equation in the standard form \( \frac{dy}{dx} + Py = Q \) where \( P = -1 \) and \( Q = x \).
3. The integrating factor (IF), \( \mu(x) \), is given by \( e^{\int P\,dx} = e^{\int -1\,dx} = e^{-x} \).
4. Multiplying the entire differential equation by the integrating factor, we have: \[ e^{-x}\frac{dy}{dx} - e^{-x}y = xe^{-x} \]
5. This equation can be rewritten as: \[ \frac{d}{dx}(ye^{-x}) = xe^{-x} \]
6. Integrate both sides with respect to \( x \): \[ ye^{-x} = \int xe^{-x} dx \]
7. The integration on the right-hand side can be done using integration by parts, where: \[ \int xe^{-x} dx = -xe^{-x} - \int (-e^{-x}) dx = -xe^{-x} + e^{-x} + C \]
8. Thus, \[ ye^{-x} = -xe^{-x} + e^{-x} + C \]
9. Multiplying throughout by \( e^{x} \) to solve for \( y \), we have: \[ y = -x + 1 + Ce^{x} \]
10. Now, apply the initial conditions to find the particular solutions:
    • For \( y_1(x) \), given \( y_1(0) = 0 \): \[ 0 = -(0) + 1 + Ce^{0} \Rightarrow C = -1 \] Hence, \( y_1(x) = -x + 1 - e^{x} \).
    • For \( y_2(x) \), given \( y_2(0) = 1 \): \[ 1 = -(0) + 1 + Ce^{0} \Rightarrow C = 0 \] Hence, \( y_2(x) = -x + 1 \).
11. To find the points of intersection, solve \( y_1(x) = y_2(x) \): \[ -x + 1 - e^{x} = -x + 1 \Rightarrow -e^{x} = 0 \]
12. As \( e^{x} \neq 0 \) for any real \( x \), there are no solutions, meaning the graphs of \( y_1(x) \) and \( y_2(x) \) do not intersect.

Therefore, the number of points of intersection is 0. The correct answer is 0.

Answer 2

The correct answer is option (A):

\[ \frac{dy}{dx} = x + y \]

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Step 1: Substitution

Let \[ x + y = t \]

Differentiating both sides with respect to x: \[ \frac{dt}{dx} = 1 + \frac{dy}{dx} \]

Using the given differential equation: \[ \frac{dy}{dx} = t \]

Therefore, \[ \frac{dt}{dx} = 1 + t \]

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Step 2: Solve the Differential Equation

Rearranging: \[ \frac{dt}{dx} - 1 = t \]

\[ \frac{dt}{t + 1} = dx \]

Integrating both sides: \[ \int \frac{dt}{t + 1} = \int dx \]

\[ \ln |t + 1| = x + C \]

\[ |t + 1| = Ce^x \]

Substituting back \( t = x + y \): \[ |x + y + 1| = Ce^x \]

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Step 3: Find Particular Solutions

First solution:

Given \( y_1(0) = 0 \)

\[ |0 + 0 + 1| = C \]

\[ C = 1 \]

Hence, \[ |x + y_1 + 1| = e^x \]

 

Second solution:

Given \( y_2(0) = 1 \)

\[ |0 + 1 + 1| = C \]

\[ C = 2 \]

Hence, \[ |x + y_2 + 1| = 2e^x \]

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Step 4: Point of Intersection

At the point of intersection:

\[ e^x = 2e^x \]

This equation has no solution.

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Final Conclusion

Therefore, the two solutions \( y_1(x) \) and \( y_2(x) \) do not intersect.