Solve the initial value problem.
-$$\frac { d y } { d x } = x \cos \left( 8 x ^ { 2 } \right) , \quad y ( 0 ) = 8$$
A) $y = \sin \left( 8 x ^ { 2 } \right) + 8$
B) $\mathrm { y } = \frac { 1 } { \mathrm { u } } \sin ( \mathrm { u } )$
C) $y = \frac { 1 } { 16 } \sin \left( 8 x ^ { 2 } \right) + 8$
D) $y = \frac { x ^ { 2 } } { 2 } \sin \left( 8 x ^ { 2 } \right) + 8$

Question

Quizplus · Accepted Answer

We can integrate both sides with respect to $x$ to get $y(x)=\int{x\cos\left(8x^2ight)dx}$. Using substitution, let $u=8x^2$, so $du=16xdx$, and $dx=\frac{1}{16x}du$. Therefore,
$$y(x)=\int{x\cos\left(8x^2ight)dx}=\int{\frac{1}{16}\cos u du}=\frac{1}{16}\sin u+C=\frac{1}{16}\sin\left(8x^2ight)+C.$$ We can find $C$ by plugging in $x=0$ to get $y(0)=C+\frac{1}{16}\sin(0)=8.$ Hence, $C=8$ and $y(x)=\frac{1}{16}\sin\left(8x^2ight)+8$, which matches choice $\boxed{	extbf{(C)}}$.

Solve the Initial Value Problem dydx=xcos⁡(8x2),y(0)=8\frac { d y } { d x } = x \cos \left( 8 x ^ { 2 } \right) , \quad y ( 0 ) = 8dxdy​=xcos(8x2),y(0)=8

Solve the Initial Value Problem $\frac { d y } { d x } = x \cos \left( 8 x ^ { 2 } \right) , \quad y ( 0 ) = 8$