Differential Equation Problem

We are given the differential equation:

\( \left(\frac{dy}{dx}\right)^2 - \frac{d^2y}{dx^2} = e^y \)

with the boundary conditions:

• \( y(0) = 0 \)
• \( y'(0) = -1 \)

We need to find the solution \( y(x) \) that satisfies this equation and the given conditions. We will verify the proposed solution.

Solution Verification

Let's verify the proposed solution \( y(x) = -\ln \left(\frac{x^2}{2}+x+1\right) \) from the options.

Verify Boundary Condition \( y(0) = 0 \)

Substitute \( x=0 \) into the proposed solution:

\( y(0) = -\ln \left(\frac{0^2}{2}+0+1\right) \)

\( y(0) = -\ln \left(0+0+1\right) \)

\( y(0) = -\ln(1) \)

Since \( \ln(1) = 0 \), we have:

\( y(0) = -0 = 0 \)

The boundary condition \( y(0) = 0 \) is satisfied by the proposed solution.

Calculate \( dy/dx \) (\( y'(x) \))

We differentiate the proposed solution \( y(x) = -\ln \left(\frac{x^2}{2}+x+1\right) \) with respect to \( x \). Using the chain rule \( \frac{d}{dx}(\ln(f(x))) = \frac{f'(x)}{f(x)} \), we get:

\( y'(x) = - \frac{1}{\frac{x^2}{2}+x+1} \cdot \frac{d}{dx}\left(\frac{x^2}{2}+x+1\right) \)

\( y'(x) = - \frac{1}{\frac{x^2}{2}+x+1} \cdot (x+1) \)

\( y'(x) = - \frac{x+1}{\frac{x^2}{2}+x+1} \)

Verify Boundary Condition \( y'(0) = -1 \)

Substitute \( x=0 \) into the expression for \( y'(x) \):

\( y'(0) = - \frac{0+1}{\frac{0^2}{2}+0+1} \)

\( y'(0) = - \frac{1}{0+0+1} \)

\( y'(0) = - \frac{1}{1} = -1 \)

The boundary condition \( y'(0) = -1 \) is satisfied by the proposed solution.

Calculate \( d^2y/dx^2 \) (\( y''(x) \))

Now we differentiate \( y'(x) = - \frac{x+1}{\frac{x^2}{2}+x+1} \) with respect to \( x \). We use the quotient rule \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \). Let \( u = -(x+1) \) and \( v = \frac{x^2}{2}+x+1 \).

Then \( u' = \frac{d}{dx}(-(x+1)) = -1 \) and \( v' = \frac{d}{dx}\left(\frac{x^2}{2}+x+1\right) = x+1 \).

\( y''(x) = \frac{(-1)\left(\frac{x^2}{2}+x+1\right) - (-(x+1))(x+1)}{\left(\frac{x^2}{2}+x+1\right)^2} \)

\( y''(x) = \frac{-\frac{x^2}{2}-x-1 + (x+1)^2}{\left(\frac{x^2}{2}+x+1\right)^2} \)

\( y''(x) = \frac{-\frac{x^2}{2}-x-1 + (x^2+2x+1)}{\left(\frac{x^2}{2}+x+1\right)^2} \)

\( y''(x) = \frac{-\frac{x^2}{2}-x-1 + x^2+2x+1}{\left(\frac{x^2}{2}+x+1\right)^2} \)

\( y''(x) = \frac{\frac{x^2}{2}+x}{\left(\frac{x^2}{2}+x+1\right)^2} \)

Calculate \( e^y \)

From the proposed solution, \( y(x) = -\ln \left(\frac{x^2}{2}+x+1\right) \). We calculate \( e^y \):

\( e^y = e^{-\ln \left(\frac{x^2}{2}+x+1\right)} \)

Using the property \( e^{-\ln A} = e^{\ln A^{-1}} = A^{-1} \), we get:

\( e^y = \left(\frac{x^2}{2}+x+1\right)^{-1} \)

\( e^y = \frac{1}{\frac{x^2}{2}+x+1} \)

Substitute into the Differential Equation

The differential equation is \( \left(\frac{dy}{dx}\right)^2 - \frac{d^2y}{dx^2} = e^y \). We substitute the expressions for \( y'(x) \), \( y''(x) \), and \( e^y \):

Left side: \( (y')^2 - y'' = \left(-\frac{x+1}{\frac{x^2}{2}+x+1}\right)^2 - \frac{\frac{x^2}{2}+x}{\left(\frac{x^2}{2}+x+1\right)^2} \)

\( (y')^2 - y'' = \frac{(x+1)^2}{\left(\frac{x^2}{2}+x+1\right)^2} - \frac{\frac{x^2}{2}+x}{\left(\frac{x^2}{2}+x+1\right)^2} \)

\( (y')^2 - y'' = \frac{x^2+2x+1 - (\frac{x^2}{2}+x)}{\left(\frac{x^2}{2}+x+1\right)^2} \)

\( (y')^2 - y'' = \frac{x^2+2x+1 - \frac{x^2}{2}-x}{\left(\frac{x^2}{2}+x+1\right)^2} \)

\( (y')^2 - y'' = \frac{\frac{x^2}{2}+x+1}{\left(\frac{x^2}{2}+x+1\right)^2} \)

\( (y')^2 - y'' = \frac{1}{\frac{x^2}{2}+x+1} \)

Right side: \( e^y = \frac{1}{\frac{x^2}{2}+x+1} \)

Since the left side equals the right side, the differential equation is satisfied by the proposed solution.

Conclusion

Since the proposed solution \( y(x) = -\ln \left(\frac{x^2}{2}+x+1\right) \) satisfies both the differential equation \( \left(\frac{dy}{dx}\right)^2 - \frac{d^2y}{dx^2} = e^y \) and the given boundary conditions \( y(0) = 0 \) and \( y'(0) = -1 \), it is the correct solution.