7:[["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"Quiz\",\"name\":\"Quiz 1\",\"description\":\"Practice problems on Differential Equations\",\"url\":\"https://theschoolofmathematics.com/quiz/ap-calculus-bc-differential-equations-quiz-2\",\"timeRequired\":\"PT15M\",\"numberOfQuestions\":9,\"isAccessibleForFree\":true,\"about\":{\"@type\":\"Thing\",\"name\":\"Differential Equations\"}}"}}],["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"BreadcrumbList\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https://theschoolofmathematics.com\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"Quizzes\",\"item\":\"https://theschoolofmathematics.com/quiz\"},{\"@type\":\"ListItem\",\"position\":3,\"name\":\"Quiz 1\",\"item\":\"https://theschoolofmathematics.com/quiz/ap-calculus-bc-differential-equations-quiz-2\"}]}"}}],["$","div",null,{"className":"bg-gradient-to-b from-white via-gray-50 to-white","children":["$","div",null,{"className":"container mx-auto px-4 py-8 md:py-12","children":[["$","div",null,{"className":"mb-12","children":["$undefined",["$","div",null,{"className":"mb-8","children":["$","h1",null,{"className":"text-4xl md:text-5xl lg:text-6xl font-bold text-gray-900 mb-4","children":"Quiz 1"}]}],["$","div",null,{"className":"bg-white rounded-2xl shadow-lg p-6 md:p-8 border border-gray-200","children":[["$","div",null,{"className":"text-lg md:text-xl text-gray-700 mb-8 leading-relaxed prose prose-lg max-w-none","dangerouslySetInnerHTML":{"__html":"

Practice problems on Differential Equations

"}}],["$","div",null,{"className":"grid grid-cols-2 md:grid-cols-3 gap-4","children":[["$","div",null,{"className":"flex items-center gap-3 p-4 bg-gray-50 rounded-xl border border-gray-200","children":[["$","div",null,{"className":"w-10 h-10 bg-gray-900 rounded-lg flex items-center justify-center flex-shrink-0","children":["$","svg",null,{"className":"w-5 h-5 text-white","fill":"none","stroke":"currentColor","viewBox":"0 0 24 24","children":["$","path",null,{"strokeLinecap":"round","strokeLinejoin":"round","strokeWidth":2,"d":"M12 8v4l3 3m6-3a9 9 0 11-18 0 9 9 0 0118 0z"}]}]}],["$","div",null,{"children":[["$","div",null,{"className":"text-xs text-gray-500 font-medium","children":"Duration"}],["$","div",null,{"className":"text-sm font-bold text-gray-900","children":"15 min"}]]}]]}],["$","div",null,{"className":"flex items-center gap-3 p-4 bg-gray-50 rounded-xl border border-gray-200","children":[["$","div",null,{"className":"w-10 h-10 bg-gray-900 rounded-lg flex items-center justify-center flex-shrink-0","children":["$","svg",null,{"className":"w-5 h-5 text-white","fill":"none","stroke":"currentColor","viewBox":"0 0 24 24","children":["$","path",null,{"strokeLinecap":"round","strokeLinejoin":"round","strokeWidth":2,"d":"M8.228 9c.549-1.165 2.03-2 3.772-2 2.21 0 4 1.343 4 3 0 1.4-1.278 2.575-3.006 2.907-.542.104-.994.54-.994 1.093m0 3h.01M21 12a9 9 0 11-18 0 9 9 0 0118 0z"}]}]}],["$","div",null,{"children":[["$","div",null,{"className":"text-xs text-gray-500 font-medium","children":"Questions"}],["$","div",null,{"className":"text-sm font-bold text-gray-900","children":10}]]}]]}],"$undefined",["$","div",null,{"className":"flex items-center gap-3 p-4 bg-gray-50 rounded-xl border border-gray-200","children":[["$","div",null,{"className":"w-10 h-10 bg-gray-900 rounded-lg flex items-center justify-center flex-shrink-0","children":["$","svg",null,{"className":"w-5 h-5 text-white","fill":"none","stroke":"currentColor","viewBox":"0 0 24 24","children":["$","path",null,{"strokeLinecap":"round","strokeLinejoin":"round","strokeWidth":2,"d":"M7 7h.01M7 3h5c.512 0 1.024.195 1.414.586l7 7a2 2 0 010 2.828l-7 7a2 2 0 01-2.828 0l-7-7A1.994 1.994 0 013 12V7a4 4 0 014-4z"}]}]}],["$","div",null,{"children":[["$","div",null,{"className":"text-xs text-gray-500 font-medium","children":"Category"}],["$","div",null,{"className":"text-sm font-bold text-gray-900","children":"Differential Equations"}]]}]]}]]}]]}]]}],["$","$L17",null,{"quiz":{"id":2222,"prismaId":"cmpz9r66l05k7jgllkfvz3a93","slug":"ap-calculus-bc-differential-equations-quiz-2","title":{"rendered":"Quiz 1"},"content":{"rendered":"

Differential Equations Quiz 2

"},"excerpt":{"rendered":"

Practice problems on Differential Equations

"},"featured_media":0,"featured_media_url":"$undefined","acf":{"duree_estimee":15,"niveau_difficulte":"$undefined","categorie":"Differential Equations","nombre_questions":10,"score_minimum":70,"ordre_questions":"Fixe","questions":[{"id":"cmpz9r66o05k9jgllycq7tjnc","texte_question":"

Find the particular solution to $$\\frac{dy}{dx}=\\frac{\\sqrt{y}}{9+x^2}$$ that satisfies the condition $$y(0)=4$$.

","type_question":"QCM","explication":"

Separate variables: $$\\frac{dy}{\\sqrt{y}}=\\frac{dx}{9+x^2}$$ Integrate both sides: $$\\int y^{-1/2}\\,dy=\\int \\frac{dx}{9+x^2}$$ $$2\\sqrt{y}=\\frac{1}{3}\\tan^{-1}\\left(\\frac{x}{3}\\right)+C$$ Use $$y(0)=4$$: $$2\\sqrt{4}=4=C$$ So $$2\\sqrt{y}=\\frac{1}{3}\\tan^{-1}\\left(\\frac{x}{3}\\right)+4$$ $$\\sqrt{y}=2+\\frac{1}{6}\\tan^{-1}\\left(\\frac{x}{3}\\right)$$ $$y=\\left(2+\\frac{1}{6}\\tan^{-1}\\left(\\frac{x}{3}\\right)\\right)^2$$ Final answer: $$\\left(2+\\frac{1}{6}\\tan^{-1}\\left(\\frac{x}{3}\\right)\\right)^2$$ Note: the source choices used $$\\tan^{-1}(x)$$ instead of $$\\tan^{-1}(x/3)$$, so the answer set was corrected.

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\left(2+\\frac{1}{3}\\tan^{-1}\\left(\\frac{x}{3}\\right)\\right)^2$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$\\left(2+\\frac{1}{6}\\tan^{-1}\\left(\\frac{x}{3}\\right)\\right)^2$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"$$$\\left(2+\\frac{1}{2}\\tan^{-1}\\left(\\frac{x}{3}\\right)\\right)^2$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"

$$(4+\\tan^{-1}\\left(\\frac{x}{3}\\right))^2$$

","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmpz9r66r05kfjgllh9nzds9r","texte_question":"

What function has derivative $$f'(x)=e^x\\sec x$$ and passes through the origin?

","type_question":"QCM","explication":"

A function with derivative $$e^x\\sec x$$ and passing through the origin can be written as $$f(x)=\\int_0^x e^t\\sec t\\,dt$$ Then $$f(0)=0$$ and by the Fundamental Theorem of Calculus, $$f'(x)=e^x\\sec x$$ Final answer: $$f(x)=\\int_0^x e^t\\sec t\\,dt$$ Note: the source answer choices did not include the correct function, so the answer set was corrected.

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$f(x)=e^x\\sec x$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$f(x)=e^x\\tan x$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$f(x)=\\int_0^x e^t\\sec t\\,dt$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"

$$f(x)=e^x\\sec x-e^0\\sec0$$

","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmpz9r66t05kljgllvutt2ebx","texte_question":"

In some cases, the rate of change of a quantity is proportional to the quantity itself. This is written as $$\\frac{dy}{dt}=ky$$. What is the general solution?

","type_question":"QCM","explication":"

Separate variables: $$\\frac{dy}{y}=k\\,dt$$ Integrate: $$\\ln|y|=kt+C$$ Exponentiate: $$y=Ce^{kt}$$ Final answer: $$y=Ce^{kt}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$y=Cke^t$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$y=Ck^t$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$y=Ce^{t/k}$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"

$$y=Ce^{kt}$$

","correcte":true,"explication":"

Correct

","imageUrl":"$undefined"}]},{"id":"cmpz9r66w05krjgll53i94ccq","texte_question":"

A particle moves along the x-axis with acceleration at time $$t$$ given by $$a(t)=4t$$. Find the function describing the particle's position if it travels at a rate of $$1$$ ft/sec when $$t=0$$ and is $$2$$ feet to the right of the origin when $$t=1$$.

","type_question":"QCM","explication":"

Since $$a(t)=s''(t)=4t$$, integrate once: $$v(t)=s'(t)=2t^2+C$$ Use $$v(0)=1$$: $$1=C$$ So $$v(t)=2t^2+1$$ Integrate again: $$s(t)=\\frac{2}{3}t^3+t+C_2$$ Use $$s(1)=2$$: $$\\frac{2}{3}+1+C_2=2$$ $$C_2=\\frac{1}{3}$$ Thus $$s(t)=\\frac{2}{3}t^3+t+\\frac{1}{3}$$ Final answer: $$\\frac{2}{3}t^3+t+\\frac{1}{3}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$s(t)=\\frac{2}{3}t^3+t+\\frac{1}{3}$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"$$$s(t)=\\frac{2}{3}t^3+t$$","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"},{"texte":"$$$s(t)=\\frac{4}{3}t^3+t$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"

$$s(t)=\\frac{2}{3}t^3+t-\\frac{1}{3}$$

","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmpz9r66y05kxjgllib1ubf4u","texte_question":"

If a ball is thrown upward and reaches its maximum height when $$t=3$$ seconds, and if the ball is $$120$$ feet off the ground when $$t=2$$ seconds, what is the position function for the ball?

","type_question":"QCM","explication":"

For motion in feet, acceleration is $$-32$$, so the position function in vertex form is $$s(t)=-16(t-3)^2+k$$ Use $$s(2)=120$$: $$120=-16(2-3)^2+k$$ $$120=-16+k$$ $$k=136$$ So $$s(t)=-16(t-3)^2+136$$ Final answer: $$-16(t-3)^2+136$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$s(t)=-16(t-3)^2+152$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$s(t)=-16(t-3)^2+136$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"$$$s(t)=-16(t-3)^2+120$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"

$$s(t)=-32(t-3)^2+136$$

","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmpz9r67105l3jgllbatpj7mc","texte_question":"

Use Euler's Method to approximate $$y(2)$$ if $$\\frac{dy}{dx}=\\frac{x}{y}$$ and $$y(1)=2$$ using one step of size $$h=1$$.

","type_question":"QCM","explication":"

Euler's Method says $$y_{n+1}=y_n+h\\,f(x_n,y_n)$$ Here $$x_0=1,\\ y_0=2,\\ h=1,\\ f(x,y)=\\frac{x}{y}$$ So $$y(2)\\approx 2+1\\cdot\\frac{1}{2}=2.5$$ Final answer: $$2.5$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$2.25$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$3$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$2.5$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"

$$2.75$$

","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmpz9r67305l9jgllvi2me061","texte_question":"

Use Euler's Method to approximate $$y(4)$$ for $$\\frac{dy}{dx}=\\frac{2}{x}$$ given that $$y(3)=1$$ using one step.

","type_question":"QCM","explication":"

Use one Euler step from $$x=3$$ to $$x=4$$, so $$h=1$$. $$f(x,y)=\\frac{2}{x}$$ Then $$y(4)\\approx y(3)+1\\cdot\\frac{2}{3}=1+\\frac{2}{3}=\\frac{5}{3}$$ Final answer: $$\\frac{5}{3}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"

$$\\frac{7}{3}$$

","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"},{"texte":"

$$2$$

","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"},{"texte":"

$$\\frac{4}{3}$$

","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"},{"texte":"

$$\\frac{5}{3}$$

","correcte":true,"explication":"

Correct

","imageUrl":"$undefined"}]},{"id":"cmpz9r67505lfjglljn9c91ms","texte_question":"

A culture of bacteria grows at a rate proportional to the number present. Initially, there are $$20$$ bacteria. After $$10$$ minutes, the population grows to $$60$$ bacteria. How many bacteria will there be after $$30$$ minutes?

","type_question":"QCM","explication":"

Exponential growth model: $$P(t)=P_0e^{kt}$$ Given: $$P(0)=20$$ $$P(10)=60$$ Solve for $$k$$: $$60=20e^{10k}$$ $$3=e^{10k}$$ $$k=\\frac{\\ln3}{10}$$ Now find $$P(30)$$: $$P(30)=20e^{30k}$$ $$=20e^{3\\ln3}$$ $$=20\\cdot27$$ $$=540$$ Final answer: $$540$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"

$$180$$

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$540$$

","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"

$$120$$

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$360$$

","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmpz9r67805lljglllwfudlzj","texte_question":"

A radioactive isotope has a half-life of $$20$$ days. If a sample initially has a mass of $$5$$ kilograms, how long will it take for the sample to decay to $$200$$ grams?

","type_question":"QCM","explication":"

Convert units: $$5\\text{ kg}=5000\\text{ g}$$ Exponential decay: $$M(t)=M_0\\left(\\frac12\\right)^{t/20}$$ Set $$200=5000\\left(\\frac12\\right)^{t/20}$$ $$\\frac{200}{5000}=\\left(\\frac12\\right)^{t/20}$$ $$\\frac1{25}=\\left(\\frac12\\right)^{t/20}$$ Take logs: $$\\ln\\left(\\frac1{25}\\right)=\\frac{t}{20}\\ln\\left(\\frac12\\right)$$ Solve: $$t=20\\frac{\\ln(1/25)}{\\ln(1/2)}$$ $$t\\approx92.9$$ Final answer: $$92\\text{ days (approx)}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"

$$86\\text{ days}$$

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$92\\text{ days}$$

","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"

$$100\\text{ days}$$

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$110\\text{ days}$$

","correcte":false,"explication":"","imageUrl":"$undefined"}]}]},"categories":["Differential Equations"],"date":"2026-06-04T09:04:28.941Z","modified":"2026-06-04T09:04:28.941Z"}}],["$","div",null,{"className":"container mx-auto px-4 pb-12","children":["$","$L18",null,{"targetType":"quiz","targetSlug":"ap-calculus-bc-differential-equations-quiz-2"}]}]]}]}]]

Quiz 1

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