Particular Solution as a Definite Integral

Question

Let $$y=r(x)$$ be the particular solution to the differential equation $$\frac{dy}{dx}=\frac{1}{\sqrt{1+x^4}}$$ with the initial condition $$r(2)=5$$. Which of the following defines $$r(x)$$?

Accepted Answer

$$r(x)=5+\int_2^x\frac{1}{\sqrt{1+t^4}}\, dt$$

Answer

$$r(x)=\frac{-2x^3}{(1+x^4)^{3/2}}$$

Answer

$$r(x)=\arctan(x^2)+C$$

Answer

$$r(x)=5+\int_0^2\frac{1}{\sqrt{1+t^4}}\, dt$$

Rank	User	Correct Count	Attempt Count	Total Time	Score
#1	haneesh19.12	1	1	0m 00s	100
#2	eftahsam	1	1	0m 25s	75
#3	npgiahy151108	1	1	1m 16s	24
#4	p250001155	0	1	0m 00s	-10
#5	seojunj	1	2	1m 50s	-20

Particular Solution as a Definite Integral

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