91Ó°ÊÓ

We assume that the second solution can be expressed as \(y_2 = u(x) \cdot y_1\), where \(u(x)\) is an unknown function. In this case, we have: $$ y_2 = u(x) \cdot e^{2x} $$

Let's find the first and second derivatives of \(y_2\): First derivative: $$ y_2' = u'(x)e^{2x} + 2u(x)e^{2x} $$ Second derivative: $$ y_2'' = u''(x)e^{2x} + 4u'(x)e^{2x} + 4u(x)e^{2x} $$

Replace \(y\), \(y'\) and \(y''\) with their expressions as functions of \(u(x)\) in the given equation, we obtain: $$ (2x + 1)(u''(x)e^{2x} + 4u'(x)e^{2x} + 4u(x)e^{2x}) - 4(x+1)(u'(x)e^{2x} + 2u(x)e^{2x}) + 4(u(x)e^{2x}) = 0 $$ Notice that \(e^{2x}\) is a common factor in every term.

Let's cancel out \(e^{2x}\) and simplify the equation: $$ (2x + 1)(u''(x) + 4u'(x) + 4u(x)) - 4(x+1)(u'(x) + 2u(x)) + 4u(x) = 0 $$ Now, let's expand, group similar terms and simplify: $$ 2xu''(x) + 4u''(x) - 4u'(x) -8u'(x) - 4u(x) + 4u(x) = 0 $$ After simplification, we get: $$ u''(x) + 2u'(x) = 0 $$ This is a first-order linear homogeneous differential equation for \(u'(x)\).

Let \(v(x) = u'(x)\): $$ v'(x) + 2v(x) = 0 $$ This can be solved using an integrating factor. The integrating factor is given by the exponential of the integral of the coefficient of \(v(x)\): $$ IF = e^{\int 2 dx} = e^{2x} $$ Now, multiply through the equation by the integrating factor: $$ e^{2x}v'(x) + 2e^{2x}v(x) = 0 $$ This can be rewritten as: $$ \frac{d}{dx}(ve^{2x}) = 0 $$ Now, integrate both sides with respect to \(x\): $$ ve^{2x} = C $$ We can now find \(v(x)\): $$ v(x) = u'(x) = \frac{C}{e^{2x}} $$ Now, we integrate \(u'(x)\) to get \(u(x)\): $$ u(x) = -\frac{C}{2e^{2x}} + D $$ To get \(y_2\), we multiply \(u(x)\) by \(y_1\): $$ y_2 = u(x)e^{2x} = -\frac{C}{2} + De^{2x} $$ However, since \(y_1 = e^{2x}\) is a solution, we can take \(y_2 = -\frac{C}{2} = const.\), giving a linearly independent solution from \(y_1\).

The general solution is a linear combination of \(y_1\) and \(y_2\): $$ y(x) = A e^{2x} + B(-\frac{1}{2}) $$ Where \(A\) and \(B\) are constants.