Solve the initial value problem.
-$$\frac { d y } { d x } = x ^ { 4 } \left( x ^ { 5 } - 5 \right) ^ { 2 } , \quad y ( 1 ) = 5$$
A) $y = \frac { 1 } { 15 } x ^ { 15 } - 5 + \frac { 8 } { 5 }$
B) $y = \frac { 1 } { 15 } \left( x ^ { 5 } - 5 \right) ^ { 3 } + \frac { 139 } { 15 }$
C) $y = \frac { 1 } { 5 } \left( x ^ { 5 } - 5 \right) ^ { 1 } + \frac { 29 } { 5 }$
D) $y = \frac { 1 } { 3 } \left( x ^ { 5 } - 5 \right) ^ { 3 } + \frac { 19 } { 15 }$

Question

Quizplus · Accepted Answer

The given differential equation is separable. We can write it as:
$$\frac{dy}{dx} = x^4(x^5-5)^2$$
Dividing both sides by $(x^5-5)^2$ and integrating, we get:
$$\int \frac{1}{(x^5-5)^2} dy = \int x^4 dx$$
Let $u=x^5-5$, then $du=5x^4dx$ and the above equation becomes:
$$\frac15 \int \frac{1}{u^2} du = \frac15 \int x^4 dx$$
This gives:
$$-\frac{1}{5(x^5-5)} = \frac{1}{15}x^5 + C$$
where $C$ is the constant of integration. Using the initial condition $y(1)=5$, we can solve for $C$:
$$-\frac{1}{5(1^5-5)} = \frac{1}{15} + C \Rightarrow C=\frac{8}{5}$$
Substituting this back into the equation above, we obtain:
$$-\frac{1}{5(x^5-5)} = \frac{1}{15}x^5 + \frac{8}{5}$$
Multiplying both sides by $(x^5-5)^2$, we get:
$$-\frac{(x^5-5)^2}{5} = \frac{1}{15}x^{10} + \frac{8}{5}(x^5-5)$$
Finally, solving for $y$ gives:
$$y = \frac{1}{15}(x^5-5)^3 + \frac{139}{15}$$
Therefore, the answer is B.

Solve the Initial Value Problem dydx=x4(x5−5)2,y(1)=5\frac { d y } { d x } = x ^ { 4 } \left( x ^ { 5 } - 5 \right) ^ { 2 } , \quad y ( 1 ) = 5dxdy​=x4(x5−5)2,y(1)=5

Solve the Initial Value Problem $\frac { d y } { d x } = x ^ { 4 } \left( x ^ { 5 } - 5 \right) ^ { 2 } , \quad y ( 1 ) = 5$