Question 2. Consider the differential equation below, in which \( a, b, c, k \) are real constants and \( c^{2}+k^{2}>0 \) : \[ \frac{d y}{d x}=\frac{a x+b y}{c x+k y} \] (a) Find 4 specific numbers \( a, b, c, k \) for which the function \( y=x^{2}-3 x \) provides a solution for equation \( () \) on the interval \( (0,+\infty) \). Briefly explain your method. (b) Is there only one correct answer to part (a)? If so, explain why; if not, describe the set of all choices for \( (a, b, c, k) \) that should be accepted. (c) Consider the ODE () with the constants found in part (a). Find a solution satisfying \( y(2)=-\frac{49}{25} \). Hint: Start by postulating \( y=x u(x) \) for some new unknown function \( u \). Find \( u \), then recover \( y \).

GPIH1N The Asker · Advanced Mathematics

Transcribed Image Text: 2. Consider the differential equation below, in which \( a, b, c, k \) are real constants and \( c^{2}+k^{2}>0 \) : \[ \frac{d y}{d x}=\frac{a x+b y}{c x+k y} \] (a) Find 4 specific numbers \( a, b, c, k \) for which the function \( y=x^{2}-3 x \) provides a solution for equation \( (*) \) on the interval \( (0,+\infty) \). Briefly explain your method. (b) Is there only one correct answer to part (a)? If so, explain why; if not, describe the set of all choices for \( (a, b, c, k) \) that should be accepted. (c) Consider the ODE (*) with the constants found in part (a). Find a solution satisfying \( y(2)=-\frac{49}{25} \). Hint: Start by postulating \( y=x u(x) \) for some new unknown function \( u \). Find \( u \), then recover \( y \).

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【General guidance】The answer provided below has been developed in a clear step by step manner.Step1/3(a)To find values of a, b, c, and k for which the function\( \mathrm{{y}={x}^{{2}}-{3}{x}} \) is a solution to the given differential equation on the interval (0, +∞), Let us differentiate y with respect to x \( \mathrm{{y}={x}^{{2}}-{3}{x},} \) \( \mathrm{{y}'={2}{x}-{3}} \)Now substitute it into the given differential equation. We can for the values of a, b, c, and k On substituting the above equation into the given differential equation, we obtain as below\( \mathrm{{\left({2}{x}-{3}\right)}=\frac{{{a}{x}+{b}{\left({x}^{{2}}-{3}{x}\right)}}}{{{c}{x}+{k}{\left({x}^{{2}}-{3}{x}\right)}}}} \)Multiply both the sides by \( \mathrm{{c}{x}+{k}{\left({x}^{{2}}-{3}{x}\right)}} \)\( \mathrm{{\left({2}{x}-{3}\right)}{\left({c}{x}+{k}{\left({x}^{{2}}-{3}{x}\right)}\right)}={a}{x}{\left({c}{x}+{k}{\left({x}^{{2}}-{3}{x}\right)}\right)}+{b}{\left({x}^{{2}}-{3}{x}\right)}{\left({c}{x}+{k}{\left({x}^{{2}}-{3}{x}\right)}\right)}} \)Equation the similar terms on both the sides we get\( \mathrm{{\left({2}{c}-{9}{k}\right)}{x}^{{3}}+{\left({k}{c}-{6}{b}\right)}{x}^{{2}}+{\left({a}{k}-{3}{c}\right)}{x}+{b}{k}={0}} \)The above equation is true for all the values of x in the interval (0, +∞) and hence the coefficients of all the powers of x will be zero. Explanation:Please refer to solution in this step.Step2/3We get these equations\( \mathrm{{2}{c}-{9}{k}={0}} \)\( \mathrm{{k}{c}-{6}{b}={0}} \)\( \mathrm{{a}{k}-{3}{c}={0}} \)\( \mathrm{{b}{k}={0}} \)From the fourth equation it ... See the full answer