Solve the given initial value problem for y = f (x). $$\frac { d y } { d x } = \frac { x + 5 } { x ^ { 2 } + 10 x + 3 }$$ where y = 6 when x = 1
A) $y = \frac { 1 } { 2 } \ln \left| x ^ { 2 } + 10 x + 3 \right| - \frac { 1 } { 2 } \ln 14 + 6$
B) $y = \frac { 1 } { 2 } \ln \left| x ^ { 2 } + 10 x + 3 \right| - \frac { 1 } { 2 } \ln 14$
C) $y = \frac { 1 } { 2 } \ln \left| x ^ { 2 } + 10 x + 3 \right| + 6$
D) $y = \frac { 1 } { 2 } \ln \left| x ^ { 2 } + 10 x + 3 \right|$

Question

Quizplus · Accepted Answer

Integrating the given differential equation $\frac{dy}{dx} = \frac{x + 5}{x^2 + 10x + 3}$, we get $y = \frac{1}{2} \ln|x^2 + 10x + 3| + C$. Using the initial condition $y(1) = 6$, we find $C = -\frac{1}{2} \ln 14 + 6$, leading to the solution $y = \frac{1}{2} \ln|x^2 + 10x + 3| - \frac{1}{2} \ln 14 + 6$.

Solve the Given Initial Value Problem for Y = F dydx=x+5x2+10x+3\frac { d y } { d x } = \frac { x + 5 } { x ^ { 2 } + 10 x + 3 }dxdy​=x2+10x+3x+5​

Solve the Given Initial Value Problem for Y = F $\frac { d y } { d x } = \frac { x + 5 } { x ^ { 2 } + 10 x + 3 }$