Question

7. a) Solve the following equations:

(ii) (D^2 - 2DD' + D'^2) z = tan (y+x)

b) Solve: z (p-q) = z^2 + (x+y^2)

Answer 1

Answer on Question #50901 – Math – Differential Calculus | Equations

a) Solve the following equation:

$$\left(D ^ {2} - 2 D D ^ {\prime} + D ^ {\prime 2}\right) z = \tan (y + x)$$

b) Solve: $z(p - q) = z^2 + (x + y)^2$

Solution

a) $(D^{2} - 2DD^{\prime} + D^{\prime 2})z = \tan (y + x)$ or $(D - D^{\prime})^{2}z = \tan (y + x)$

Here auxiliary equation is $(m - 1)^2 = 0$ so that $m = 1$ .

Thus

$C.F. = \varphi_{1}(y + x) + x\varphi_{2}(y + x)$ , where $\varphi_{1}$ and $\varphi_{2}$ are arbitrary functions.

Now

$$P. I. = \frac {1}{(D - D ^ {\prime}) ^ {2}} \tan (y + x) = \frac {x ^ {2}}{1 ^ {2} 2 !} \tan (y + x) = \frac {x ^ {2}}{2} \tan (y + x).$$

The general solution is

$$z = \varphi_ {1} (y + x) + x \varphi_ {2} (y + x) + \frac {x ^ {2}}{2} \tan (y + x).$$

b) The auxiliary equations are

$$\frac {d x}{z} = \frac {d y}{- z} = \frac {d z}{z ^ {2} + (x + y)}.$$

Taking the first two members we get

$$- z d x = z d y \rightarrow x + y = C _ {1}.$$

Thus

$$d x (z ^ {2} + C _ {1}) = z d z \rightarrow 2 d x = \frac {d (z ^ {2} + C _ {1})}{(z ^ {2} + C _ {1})} = d (\ln (z ^ {2} + C _ {1})) \rightarrow x - \ln (\sqrt {z ^ {2} + C _ {1}}) = C _ {2}.$$ $$z = \varphi_ {1} (y + x) + \varphi_ {2} \left(x - \ln (\sqrt {z ^ {2} + x + y})\right).$$

Where $\varphi_{1}$ and $\varphi_{2}$ are arbitrary functions.

www.AssignmentExpert.com