Question 73 of 96intermediate🔧 ApplyLong Answer4 marks

For each of the differential equations given in Exercises 1 to 12, find the general solution:

(x+y)dydx=1(x + y) \frac{dy}{dx} = 1
Correct Answer

The general solution is:

x+y+1=Cey\mathbf{x + y + 1 = Ce^{y}}

(where CC is an arbitrary constant).

Exercise: EXERCISE 9.5 | Q: 10 | (Page 30)
For More Understanding

Explanation

This question is from Exercise 9.4, Question 10. The differential equation

(x+y)dydx=1(x + y) \frac{dy}{dx} = 1

can be solved by rewriting it as

dxdy=x+y\frac{dx}{dy} = x + y

which becomes a linear differential equation in xx as a function of yy. This is a standard technique where we treat xx as the dependent variable and yy as the independent variable. The equation then takes the form

dxdy−x=y\frac{dx}{dy} - x = y

which can be solved using the integrating factor method.

Solution Steps

  1. Step 1: Rewrite the given differential equation

    (x+y)dydx=1(x + y) \frac{dy}{dx} = 1

    as

    dxdy=x+y\frac{dx}{dy} = x + y

  2. Step 2: Rearrange to get the linear form:

    dxdy−x=y\frac{dx}{dy} - x = y

  3. Step 3: Find the integrating factor:

    IF=e∫−1 dy=e−yIF = e^{\int -1 \, dy} = e^{-y}

  4. Step 4: Multiply both sides by IFIF:

    e−ydxdy−e−yx=ye−ye^{-y} \frac{dx}{dy} - e^{-y}x = ye^{-y}

  5. Step 5: This simplifies to:

    ddy(xe−y)=ye−y\frac{d}{dy}(xe^{-y}) = ye^{-y}

  6. Step 6: Integrate both sides (using integration by parts):

    xe−y=∫ye−y dy=−ye−y−e−y+Cxe^{-y} = \int ye^{-y} \, dy = -ye^{-y} - e^{-y} + C

  7. Step 7: Simplify:

    x=−y−1+Ceyx = -y - 1 + Ce^{y}

  8. Step 8: The general solution is:

    x+y+1=Cey\mathbf{x + y + 1 = Ce^{y}}
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