Question 90 of 96intermediate🔧 ApplyLong Answer4 marks

Find a particular solution of the differential equation (x−y)(dx+dy)=dx−dy(x - y)(dx + dy) = dx - dy, given that y=−1y = -1, when x=0x = 0. (Hint: put x−y=tx - y = t)

Correct Answer

The particular solution is:

x+y+1=ln⁡(x−y)x + y + 1 = \ln(x - y)

or equivalently:

ex+y+1=x−ye^{x + y + 1} = x - y
Exercise: Miscellaneous Exercise on Chapter 9 | Q: 9 | (Page 36)
For More Understanding

Explanation

This question requires solving a differential equation using substitution. The hint suggests putting x−y=tx - y = t, which simplifies the equation. After substitution and integration, we apply the initial condition y=−1y = -1 when x=0x = 0 to find the constant of integration.

The solution follows standard techniques shown in Example 6 of the context where particular solutions are found by applying initial conditions.

Solution Steps

  1. Step 1: Let x−y=tx - y = t. Differentiating, we get:

    dx−dy=dtdx - dy = dt

  2. Step 2: Substitute in the given equation (x−y)(dx+dy)=dx−dy(x - y)(dx + dy) = dx - dy. We get:

    t(dx+dy)=dtt(dx + dy) = dt

    which gives:

    dx+dy=dttdx + dy = \frac{dt}{t}

  3. Step 3: Integrating both sides:

    ∫(dx+dy)=∫dtt\int(dx + dy) = \int \frac{dt}{t}

    which gives:

    x+y=ln⁡∣t∣+Cx + y = \ln|t| + C

  4. Step 4: Substitute back t=x−yt = x - y:

    x+y=ln⁡∣x−y∣+Cx + y = \ln|x - y| + C

  5. Step 5: Apply the initial condition y=−1y = -1 when x=0x = 0. At this point:

    t=x−y=0−(−1)=1t = x - y = 0 - (-1) = 1

    Substituting:

    0+(−1)=ln⁡∣1∣+C0 + (-1) = \ln|1| + C

    giving:

    −1=0+C-1 = 0 + C

    so:

    C=−1C = -1

  6. Step 6: The particular solution is:

    x+y=ln⁡∣x−y∣−1x + y = \ln|x - y| - 1

    or:

    x+y+1=ln⁡∣x−y∣x + y + 1 = \ln|x - y|

    Since x−y=1>0x - y = 1 > 0 at the initial point, we write:

    x+y+1=ln⁡(x−y)x + y + 1 = \ln(x - y)

    or equivalently:

    ex+y+1=x−ye^{x + y + 1} = x - y
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