Worksheet: Initial Value Problems

In this worksheet, we will practice finding a specific solution to a separable differential equation given an initial value.

Q1:

Find the equation of the curve that passes through the point (βˆ’3,2) given that the gradient of the tangent at any point is βˆ’4π‘₯7𝑦.

  • A7𝑦=βˆ’4π‘₯+79
  • B7𝑦=βˆ’4π‘₯+C
  • C7𝑦=βˆ’4π‘₯+792
  • D7𝑦=βˆ’4π‘₯+64

Q2:

Find the equation of the curve that passes through the point (0,βˆ’1) given dd𝑦π‘₯=βˆ’6π‘₯βˆ’44𝑦+13.

  • A2𝑦+13𝑦=βˆ’3π‘₯βˆ’4π‘₯βˆ’11
  • B2𝑦+13𝑦=βˆ’3π‘₯βˆ’4π‘₯βˆ’9
  • C4𝑦+13𝑦=βˆ’6π‘₯βˆ’4π‘₯βˆ’9
  • D2𝑦+13𝑦=βˆ’6π‘₯βˆ’4π‘₯βˆ’9
  • E4𝑦+13𝑦=βˆ’3π‘₯βˆ’4π‘₯βˆ’11

Q3:

Find the solution of the differential equation dd𝑦π‘₯=π‘₯π‘’ο˜ that satisfies the initial condition 𝑦(0)=0.

  • A𝑦=βˆ’ο€Όβˆ’12π‘₯ln
  • B𝑦=βˆ’ο€Ήβˆ’π‘₯+1ln
  • C𝑦=βˆ’ο€Ήβˆ’π‘₯ln
  • D𝑦=βˆ’ο€Όβˆ’12π‘₯+1ln
  • E𝑦=βˆ’ο€Ό12π‘₯+1ln

Q4:

Find the solution of the differential equation dd𝑃𝑑=βˆšπ‘ƒπ‘‘ that satisfies the initial condition 𝑃(1)=2.

  • Aβˆšπ‘ƒ=13π‘‘βˆ’13+√2
  • Bβˆšπ‘ƒ=13π‘‘βˆ’34+√2
  • Cβˆšπ‘ƒ=βˆ’13π‘‘βˆ’βˆš2+13
  • Dβˆšπ‘ƒ=13𝑑+13+√2
  • Eβˆšπ‘ƒ=13π‘‘βˆ’βˆš2βˆ’13

Q5:

Suppose that ddcossin𝑦π‘₯=4π‘₯βˆ’42π‘₯4𝑦+9 and 𝑦=0 when π‘₯=0. Find 𝑦 in terms of π‘₯.

  • A9π‘¦βˆ’4𝑦=4π‘₯βˆ’22π‘₯βˆ’4cossin
  • B9π‘¦βˆ’4𝑦=2π‘₯βˆ’22π‘₯βˆ’4cossin
  • C9𝑦+4𝑦=2π‘₯+22π‘₯+4cossin
  • D9𝑦+4𝑦=4π‘₯+22π‘₯+4cossin
  • E9π‘¦βˆ’4𝑦=2π‘₯βˆ’42π‘₯βˆ’4cossin

Q6:

A relation 𝑓(π‘₯,𝑦)=0 is implicitly differentiated to obtain dd𝑦π‘₯=2π‘₯+52𝑦+5. Find the relation given that when 𝑦=3, π‘₯=3.

  • Aπ‘₯+5π‘₯βˆ’2π‘¦βˆ’5𝑦=0
  • Bπ‘₯βˆ’5π‘¦βˆ’3=0
  • Cπ‘₯+5π‘₯βˆ’π‘¦βˆ’5𝑦=0
  • Dπ‘₯+5π‘₯+5π‘¦βˆ’9=0

Q7:

Find the equation of the curve that passes through the point (βˆ’8,1) given that the gradient of the tangent at any point is equal to 2 times the square of the 𝑦 coordinate.

  • A𝑦=βˆ’12π‘₯βˆ’17
  • B𝑦=βˆ’12π‘₯+15
  • C𝑦=12π‘₯βˆ’15
  • D𝑦=βˆ’12π‘₯+17

Q8:

Suppose that ddsincos𝑦π‘₯=3π‘₯42π‘¦οŠ¨οŠ¨ and 𝑦=πœ‹4 when π‘₯=πœ‹2. Find 𝑦 in terms of π‘₯.

  • A8𝑦+84𝑦=6π‘₯βˆ’62π‘₯βˆ’πœ‹sinsin
  • B8π‘¦βˆ’24𝑦=6π‘₯+32π‘₯βˆ’πœ‹sinsin
  • C8𝑦+24𝑦=6π‘₯βˆ’32π‘₯βˆ’πœ‹sinsin
  • D8π‘¦βˆ’84𝑦=6π‘₯+62π‘₯βˆ’πœ‹sinsin

Q9:

Find the solution of the differential equation dd𝑦π‘₯+9𝑦=63 given that 𝑦(0)=8.

  • A𝑦=7+π‘’οŠ―ο—
  • B𝑦=7+π‘’οŠ±ο—
  • C𝑦=7+π‘’ο‘οŽ¨
  • D𝑦=7+π‘’οŠ±οŠ―ο—

Q10:

Find the solution for the following differential equation for 𝑦(0)=2: dd𝑦π‘₯=π‘₯𝑦.οŠͺ

  • A𝑦=ο€Ή15π‘₯+8ο…οŠ«οŽ οŽ’
  • B𝑦=35π‘₯+8
  • C𝑦=ο€Ό35π‘₯+8
  • D𝑦=15π‘₯+8

Q11:

Find the solution of the differential equation ddsin𝑦π‘₯=π‘₯π‘₯𝑦 that satisfies the initial condition 𝑦(0)=βˆ’6.

  • A𝑦=2(π‘₯βˆ’π‘₯π‘₯)βˆ’36sincos
  • B𝑦=(π‘₯βˆ’π‘₯π‘₯)+36sincos
  • C𝑦=2(π‘₯βˆ’π‘₯π‘₯)+36sincos
  • D𝑦=2(π‘₯βˆ’π‘₯π‘₯)sincos
  • E𝑦=2(π‘₯π‘₯βˆ’π‘₯)+36sincos

Q12:

Find the solution of the differential equation π‘₯π‘₯=𝑦1+√1+3𝑦𝑦′ln that satisfies the initial condition 𝑦(1)=βˆ’1.

  • A12π‘₯π‘₯+14π‘₯=19ο€Ή3𝑦+1+12π‘¦οŠ¨οŠ¨οŠ¨οŠ¨ln
  • B12π‘₯π‘₯βˆ’14π‘₯+5936=23ο€Ή3𝑦+1+12π‘¦οŠ¨οŠ¨οŠ¨οŠ¨ln
  • C12π‘₯π‘₯+14π‘₯+4136=19ο€Ή3𝑦+1+12π‘¦οŠ¨οŠ¨οŠ¨οŠ¨ln
  • D12π‘₯π‘₯βˆ’14π‘₯+5936=19ο€Ή3𝑦+1+12π‘¦οŠ¨οŠ¨οŠ¨οŠ¨ln
  • E12π‘₯π‘₯βˆ’14π‘₯=19ο€Ή3𝑦+1+12π‘¦οŠ¨οŠ¨οŠ¨οŠ¨ln

Q13:

Find the solution for the following differential equation for 𝑦(2)=1: 𝑦𝑦+π‘₯π‘₯=3π‘₯𝑦π‘₯.ddd

  • A𝑦=ο„ž1+2𝑒3οŠ©ο—οŠ±οŠ§οŠ¨οŽ‘
  • B𝑦=ο„ž1+2𝑒3οŠ±οŠ©ο—οŠ±οŠ§οŠ¨
  • C𝑦=ο„ž1+2𝑒3οŠ±οŠ©ο—οŠ±οŠ§οŠ¨οŽ‘
  • D𝑦=ο„ž1+2𝑒3οŠ©ο—οŠ±οŠ§οŠ¨

Q14:

Find the solution of the differential equation π‘₯+π‘¦βˆšπ‘₯+3𝑦π‘₯=0dd that satisfies the initial condition 𝑦(βˆ’1)=βˆ’2.

  • A𝑦=βˆ’ο„3√π‘₯+3
  • B𝑦=βˆ’ο„3√π‘₯+3+2
  • C𝑦=βˆ’ο„βˆšπ‘₯+3+6
  • D𝑦=βˆ’βˆšπ‘₯+3οŽ₯
  • E𝑦=βˆ’ο„3√π‘₯+3+14

Q15:

Find the solution of the differential equation 12𝑦π‘₯+𝑦(𝑒)=0ddοŠ¨ο— that passes through the point (0,4).

  • A𝑦=12𝑒+3
  • B𝑦=1π‘’βˆ’2
  • C𝑦=12π‘’βˆ’8
  • D𝑦=12𝑒+2
  • E𝑦=1𝑒+2

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