SEPARABLE DIFFERENTIAL EQUATIONS 1

These are questions on separable differential equations

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1. Kuta Software - Infinite Calculus Name___________________________________ Separable Differential Equations Date________________ Period____ Find the general solution of each differential equation. 1) dy x− y =e dx 2) 1 dy = dx sec 2 y 3) dy y = xe dx 4) dy 2 x = dx e 2 y 5) dy = 2y − 1 dx 6) dy = 2 yx + yx 2 dx ©F D2f0B1I3F sKnuwtqay RSAo9fjt3wdaircei TLgL4CK.w j EAMlilW mrOidgxhTtysO nr3eFsGefrpv0eadO.q T 9MdaFdQe5 UwDibtuh8 UI0nMf6i3nZiitxez sCjaAljcDuClguEsb.k -1- Worksheet by Kuta Software LLC
2. For each problem, find the particular solution of the differential equation that satisfies the initial condition. You may use a graphing calculator to sketch the solution on the provided graph. dy x− y = 2e , y(1) = ln (2e + 1) dx 7) 2 dy = xy 2 , y(2) = − dx 5 8) y −4 9) −3 −2 y 4 4 3 3 2 2 1 1 −1 1 2 3 4 x −4 −3 −2 −1 1 −1 −1 −2 −2 −3 −3 −4 −4 dy = 12 x 3 y, y(0) = 2 dx 10) −3 −2 4 x 3 4 x y 4 4 3 3 2 2 1 1 −1 3 x dy = − , y(1) = − 2 dx y y −4 2 1 2 3 4 x −4 −3 −2 −1 1 −1 −1 −2 −2 −3 −3 −4 −4 ©Q O290A103x KKPubtSar BSwoKfetHwZalrIeN GLMLxC7.j W RALlMlW Wr2iNgXhZtCsU qrUeqsQearTvneadb.W M 7Mcadd8eh DwtiUtyh5 2I8npfbibnHiftOee uCPazlcc1uClTu6sQ.3 -2- 2 Worksheet by Kuta Software LLC
3. Kuta Software - Infinite Calculus Name___________________________________ Separable Differential Equations Date________________ Period____ Find the general solution of each differential equation. 1) dy x− y =e dx y 2) x tan y = x + C y = tan −1 ( x + C) e =e +C x y = ln (e + C) 3) dy y = xe dx 4) x2 −e = + C1 2 x2 y = − ln − +C 2 5) dy = 2y − 1 dx ln 2 y − 1 dy 2 x = dx e 2 y 2y e 2 = x + C1 2 ln (2 x 2 + C) y= 2 −y ( 1 dy = dx sec 2 y ) 6) ln y = x 2 + = x + C1 2 2x Ce + 1 y= 2 ©t v2w0B103m PKVuwtgaJ iSPo0fktWw0aerXeJ MLMLuCw.W O 1AilSlG Drni5gbhMtvsa FrsetsvekrYv4eIdL.0 0 FMNamdUec ewviVtIhS fIEnoftiunDiGtDeJ qCZaNldcyunlZuNsV.L dy = 2 yx + yx 2 dx x2 + y = Ce -1- x3 + C1 3 x3 3 Worksheet by Kuta Software LLC
4. For each problem, find the particular solution of the differential equation that satisfies the initial condition. You may use a graphing calculator to sketch the solution on the provided graph. dy x− y = 2e , y(1) = ln (2e + 1) dx 7) 2 dy = xy 2 , y(2) = − dx 5 8) y −4 −3 −2 y y 4 4 3 3 2 2 1 1 −1 1 2 3 4 x −4 −2 −1 1 −1 −1 −2 −2 −3 −3 −4 −4 dy = 12 x 3 y, y(0) = 2 dx 10) −3 −2 4 x 3 4 x y 4 4 3 3 2 2 1 1 −1 3 x dy = − , y(1) = − 2 dx y y −4 2 1 x2 1 − = + y 2 2 2 y=− 2 x +1 x e = 2e + 1 x y = ln (2e + 1) 9) −3 1 2 3 4 x −4 −3 −2 −1 1 −1 −1 −2 −2 −3 −3 −4 −4 ln y = 3 x 4 + ln 2 3x4 y = 2e 2 2 y x2 3 =− + 2 2 2 2 y = − −x + 3 , − 3 < x < 3 Create your own worksheets like this one with Infinite Calculus. Free trial available at KutaSoftware.com ©p r2Y0I1W3H kKQuit1aQ fSooifZtFwTaarzeA nLnLtCL.w j vA5lrlf qrpiegshZtmsd brDe1sTewrKvEeMdz.7 S yMzaIdpek ywNidtihB xIknMf9iCnyi7tWeW 1CrajlscourlFuYsE.N -2- Worksheet by Kuta Software LLC