Question 88 of 96intermediate🔧 ApplyLong Answer4 marks

Find the particular solution of the differential equation (1+e2x) dy+(1+y2) ex dx=0(1 + e^{2x})\, dy + (1 + y^2)\, e^x\, dx = 0, given that y=1y = 1 when x=0x = 0.

Correct Answer

The given differential equation is:

(1+e2x) dy+(1+y2) ex dx=0(1 + e^{2x})\, dy + (1 + y^2)\, e^x\, dx = 0

To solve, we first separate the variables. Rearranging the terms, we get:

dy1+y2=−ex1+e2x dx\frac{dy}{1 + y^2} = \frac{-e^x}{1 + e^{2x}}\, dx

Integrating both sides, we have:

∫dy1+y2=−∫ex1+e2x dx\int \frac{dy}{1 + y^2} = - \int \frac{e^x}{1 + e^{2x}}\, dx

Using the standard integral form from Example 5 for the left side, ∫dy1+y2=tan⁡−1y\int \frac{dy}{1 + y^2} = \tan^{-1} y. For the right side, we use the substitution method. Let ex=te^x = t, which implies ex dx=dte^x\, dx = dt. The integral becomes:

−∫dt1+t2=−tan⁡−1t=−tan⁡−1(ex)- \int \frac{dt}{1 + t^2} = -\tan^{-1} t = -\tan^{-1} (e^x)

Thus, the general solution is:

tan⁡−1y=−tan⁡−1(ex)+C\tan^{-1} y = -\tan^{-1} (e^x) + C

Or,

tan⁡−1y+tan⁡−1(ex)=C\tan^{-1} y + \tan^{-1} (e^x) = C

We now find the particular solution given that y=1y = 1 when x=0x = 0. Substituting these values:

tan⁡−1(1)+tan⁡−1(e0)=C\tan^{-1} (1) + \tan^{-1} (e^0) = C π4+tan⁡−1(1)=C\frac{\pi}{4} + \tan^{-1} (1) = C π4+π4=C\frac{\pi}{4} + \frac{\pi}{4} = C C=π2C = \frac{\pi}{2}

Substituting CC back into the general solution:

tan⁡−1y+tan⁡−1(ex)=π2\tan^{-1} y + \tan^{-1} (e^x) = \frac{\pi}{2}

This can be simplified as:

tan⁡−1y=π2−tan⁡−1(ex)=cot⁡−1(ex)\tan^{-1} y = \frac{\pi}{2} - \tan^{-1} (e^x) = \cot^{-1} (e^x) y=tan⁡(cot⁡−1(ex))=1ex=e−xy = \tan(\cot^{-1} (e^x)) = \frac{1}{e^x} = e^{-x}

∴ The particular solution is y=e−x\boxed{y = e^{-x}}.

Exercise: Miscellaneous Exercise on Chapter 9 | Q: 7 | (Page 34)
For More Understanding

Explanation

The solution follows the method of separation of variables demonstrated in Example 5 and Example 6 of the provided context. Example 5 establishes the integral ∫dy1+y2=tan⁡−1y\int \frac{dy}{1+y^2} = \tan^{-1}y.

The integration on the right side utilizes substitution (a method hinted at in Question 9), reducing ∫ex1+e2x dx\int \frac{e^x}{1+e^{2x}}\,dx to tan⁡−1(ex)\tan^{-1}(e^x).

Finally, the constant of integration CC is determined using the specific values provided, a procedure explicitly shown in Example 6.

Solution Steps

  1. Step 1: Separate the variables xx and yy to get dy1+y2=−ex1+e2x dx.\frac{dy}{1 + y^2} = \frac{-e^x}{1 + e^{2x}}\, dx.

  2. Step 2: Integrate both sides. LHS becomes tan⁡−1y\tan^{-1} y. For RHS, substitute ex=te^x = t to get −tan⁡−1(ex)-\tan^{-1} (e^x).

  3. Step 3: Form the general solution tan⁡−1y+tan⁡−1(ex)=C.\tan^{-1} y + \tan^{-1} (e^x) = C.

  4. Step 4: Substitute y=1y=1 and x=0x=0 to find C=π2.C = \frac{\pi}{2}.

  5. Step 5: Write the final particular solution tan⁡−1y+tan⁡−1(ex)=π2\tan^{-1} y + \tan^{-1} (e^x) = \frac{\pi}{2} or y=e−x.y = e^{-x}.

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