First order differential equations (integrating factor)
7
Show that an appropriate integrating factor for
$$\sqrt { x ^ { 2 } + 16 } \frac { \mathrm {~d} y } { \mathrm {~d} x } + y = x \sqrt { x ^ { 2 } + 16 }$$
is \(\frac { 1 } { 4 } x + \frac { 1 } { 4 } \sqrt { x ^ { 2 } + 16 }\) .
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Hence find the solution of the differential equation
$$\sqrt { x ^ { 2 } + 16 } \frac { \mathrm {~d} y } { \mathrm {~d} x } + y = x \sqrt { x ^ { 2 } + 16 }$$
for which \(y = 6\) when \(x = 3\).