Without solving for the undetermined coefficients, the correct form of a particular solution of the differential equation \( y^{\prime \prime}+4 y^{\prime}-2 y=x^{2}+4 \) \( y_{p}=A x+B \) \( y_{p}=c_{1} e^{a x}+c_{2} x e^{b x} \) \( y_{p}=A \) \( y_{p}=c_{1} e^{a x}+c_{2} e^{b x} \) \( y_{p}=A x^{2}+B x+C \)
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Beyond the Answer
To find a particular solution for the given differential equation \( y^{\prime \prime}+4 y^{\prime}-2 y=x^{2}+4 \), we first take a look at the right-hand side, which is a polynomial of degree 2 (specifically \( x^2 + 4 \)). For polynomial forcing functions, the typical approach is to assume a polynomial form for \( y_p \). Since the right-hand side is a quadratic polynomial, the most fitting form for the particular solution would be \( y_p = A x^{2} + B x + C \), where \( A \), \( B \), and \( C \) are constants to be determined later. The other forms listed, like exponential functions or constants, would not adequately capture the polynomial nature of the non-homogeneous term. Forcing functions often dictate the structure of the trial solutions. In this scenario, since we have a polynomial on the right, we gravitate towards a polynomial form on the left to match it effectively! Now, let's turn our curiosity towards the intricacies. Did you know that the method of undetermined coefficients relies on these clever assumptions? If the non-homogeneous term included trigonometric or exponential functions, we'd need to alter our approach accordingly! So, it’s all connected—each term influences how we structure our guess for the solution! Keep exploring, and you’ll see the patterns emerge!
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