Question 93 of 96intermediate🔧 ApplyLong Answer4 marks

Find a particular solution of the differential equation (x+1)dydx=2e−y−1(x + 1) \frac{dy}{dx} = 2 e^{-y} - 1, given that y=0y = 0 when x=0x = 0.

Correct Answer

The given differential equation is:

(x+1)dydx=2e−y−1(x + 1) \frac{dy}{dx} = 2 e^{-y} - 1

Separating the variables xx and yy, we rewrite the equation as:

dy2e−y−1=dxx+1\frac{dy}{2 e^{-y} - 1} = \frac{dx}{x + 1}

Multiplying the numerator and denominator of the LHS by eye^y:

eydy2−ey=dxx+1\frac{e^y dy}{2 - e^y} = \frac{dx}{x + 1}

Integrating both sides, we get:

∫ey2−eydy=∫1x+1dx\int \frac{e^y}{2 - e^y} dy = \int \frac{1}{x + 1} dx

Let 2−ey=t2 - e^y = t, then −eydy=dt⇒eydy=−dt-e^y dy = dt \Rightarrow e^y dy = -dt.

−∫dtt=log⁡(x+1)+C-\int \frac{dt}{t} = \log(x + 1) + C −log⁡(2−ey)=log⁡(x+1)+C-\log(2 - e^y) = \log(x + 1) + C

Given that y=0y = 0 when x=0x = 0. Substituting these values to find CC:

−log⁡(2−e0)=log⁡(0+1)+C-\log(2 - e^0) = \log(0 + 1) + C −log⁡(1)=log⁡(1)+C-\log(1) = \log(1) + C

Since log⁡(1)=0\log(1) = 0, we have C=0C = 0.

Substituting C=0C = 0 back into the equation:

−log⁡(2−ey)=log⁡(x+1)-\log(2 - e^y) = \log(x + 1) log⁡(2−ey)−1=log⁡(x+1)\log(2 - e^y)^{-1} = \log(x + 1) log⁡(12−ey)=log⁡(x+1)\log\left(\frac{1}{2 - e^y}\right) = \log(x + 1) 12−ey=x+1\frac{1}{2 - e^y} = x + 1 1=(x+1)(2−ey)1 = (x + 1)(2 - e^y) 2−ey=1x+12 - e^y = \frac{1}{x + 1} ey=2−1x+1=2x+2−1x+1=2x+1x+1e^y = 2 - \frac{1}{x + 1} = \frac{2x + 2 - 1}{x + 1} = \frac{2x + 1}{x + 1}

Taking logarithm on both sides, the particular solution is:

y=log⁡(2x+1x+1)y = \log\left(\frac{2x + 1}{x + 1}\right)
Exercise: Miscellaneous Exercise on Chapter 9 | Q: 12 | (Page 36)
For More Understanding

Explanation

The problem requires solving a first-order differential equation by separating the variables. The provided context illustrates this method in Example 20 and the snippet showing integration of dy/y2dy/y^2. The equation is manipulated into the form f(y)dy=g(x)dxf(y)dy = g(x)dx.

After integration, the arbitrary constant is determined using the initial condition y=0y=0 when x=0x=0, similar to the method shown in the provided snippet where CC is found by substituting xx and yy values. The final solution is derived by solving for yy.

Solution Steps

  1. Step 1: Rewrite the equation (x+1)dydx=2e−y−1(x + 1) \frac{dy}{dx} = 2 e^{-y} - 1 in separable form eydy2−ey=dxx+1\frac{e^y dy}{2 - e^y} = \frac{dx}{x + 1}.

  2. Step 2: Integrate both sides: LHS with substitution t=2−eyt = 2 - e^y yields −log⁡∣2−ey∣-\log|2 - e^y|, RHS yields log⁡∣x+1∣+C\log|x + 1| + C.

  3. Step 3: Apply the condition y=0y = 0 when x=0x = 0 to find the constant CC. Substitution gives C=0C = 0.

  4. Step 4: Solve the resulting equation −log⁡(2−ey)=log⁡(x+1)-\log(2 - e^y) = \log(x + 1) algebraically to isolate yy.

  5. Step 5: Final answer is obtained as y=log⁡(2x+1x+1)y = \log\left(\frac{2x + 1}{x + 1}\right).

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