Answer to Question #247659 in Calculus for JaytheCreator

Question #247659

Solve the following differential equations: (i) (𝑦 3 + 1) 𝑑𝑦 𝑑π‘₯ βˆ’ π‘₯𝑦 2 = π‘₯, (ii) π‘₯𝑦 2 𝑑𝑦 𝑑π‘₯ = π‘₯ 3 βˆ’ 𝑦 3, (iii) (1 + sin π‘₯) 𝑑𝑦 𝑑π‘₯ + 𝑦 cos π‘₯ = tan π‘₯, (iv) π‘₯ 𝑑𝑦 𝑑π‘₯ βˆ’ 5𝑦 = π‘₯ 7.


1
Expert's answer
2021-10-22T11:06:52-0400

(i)

"(y^{3}+1)\\frac{dy}{dx}=x+xy^{2}"

"(y^{3}+1)\\frac{dy}{dx}=x(1+y^{2})"

"\\frac{1+y^{3}}{1+y^{2}}dy=xdx"

Integrate on both sides

"\\int \\frac{1+y^{3}}{1+y^{2}}dy=\\int xdx"

Let "1+y^{2}" =t

2ydy=dt

ydy="\\frac{dt}{2}"

"y^{2}=(t-1)"

"\\int \\frac{1}{1+y^{2}}dy+\\int \\frac{y^{3}}{1+y^{2}}=\\int xdx"

"tan^{-1}(y)+\\frac{1}{2}\\int \\frac{(t-1)}{t}dt=\\frac{x^{2}}{2}+C"

"tan^{-1}(y)+\\frac{1}{2}\\int (1-\\frac{1}{t})dt=\\frac{x^{2}}{2}+C"

"tan^{-1}(y)+\\frac{1}{2}(t-ln\\ t)=\\frac{x^{2}}{2}+C"

Putting the values of t answer becomes;

"tan^{-1}(y)+\\frac{1}{2}[(1+y^{2})-ln(1+y^{2})]=\\frac{x^{2}}{2}+C"


(ii)

"xy^{2}\\frac{dy}{dx}=x^{3}-y^{3}"

"\\frac{dy}{dx}=\\frac{x^{3}-y^{3}}{xy^{2}}"

On dividing the numerator and denominator by x3

"\\frac{dy}{dx}=\\frac{1-(\\frac{y}{x})^{3}}{(\\frac{y}{x})^{2}}"

Let "\\frac{y}{x}=V" "\\implies y=Vx"

"\\frac{dy}{dx}=V+x\\frac{dV}{dx}"

"V+x\\frac{dV}{dx}=\\frac{1-V^{3}}{V^{2}}"

"x\\frac{dV}{dx}=\\frac{1-V^{3}}{V^{2}}-V"

"x\\frac{dV}{dx}=\\frac{1-V^{3}-V^{3}}{V^{2}}"

"x\\frac{dV}{dx}=\\frac{1-2V^{3}}{V^{2}}"

"\\frac{V^{2}}{1-2V^{3}}dV=\\frac{1}{x}dx"

Integrating both sides

"\\int \\frac{V^{2}}{1-2V^{3}}dV=\\int \\frac{1}{x}dx"

Let 1-2V3=t

-6V2dV=dt

"V^{2}dV=-\\frac{1}{6}dt"

"\\frac{-1}{6}\\int \\frac{1}{t}dt=\\int \\frac{1}{x}dx"

"-\\frac{1}{6}ln\\ t=ln\\ x+C"

"-\\frac{1}{6}ln\\ (1-2V^{3})=ln\\ x+C"

"-\\frac{1}{6}ln\\ (1-2(\\frac{y}{x})^{3})=ln\\ x+C"

"-\\frac{1}{6}ln[\\frac{x^{3}-2y^{3}}{x^{3}}]=ln\\ x+C"


(iii)

"(1+sin\\ x)\\frac{dy}{dx}+ycos\\ x=tan\\ x"

"\\frac{dy}{dx}+\\frac{ycos\\ x}{(1+sin\\ x)}=\\frac{tan\\ x}{(1+sin\\ x)}"

This is a linear equation in y

Integrating factor ,I.F="e^{\\int{\\frac{cos\\ x}{1+sin\\ x}}dx}"

"=e^{ln(1+sin\\ x)}"

"=1+sin\\ x"

Multiply through by I.F to get;

"\\frac{d}{dx}[y(1+sin\\ x)]=\\frac{tan\\ x}{1+sin\\ x}(1+sin\\ x)"

"\\frac{d}{dx}[y(1+sin\\ x)]={tan\\ x}"

Integrating;

"y(1+sin\\ x)=\\int tan\\ xdx"

"y(1+sin\\ x)=ln|sec\\ x|+C"

Our solution therefore is;

"y=\\frac{ln|sex\\ x|}{1+sin\\ x}+\\frac{C}{1+sin\\ x}"

(iv)

"x\\frac{dy}{dx}-5y=x^{7}"

"\\frac{dy}{dx}-\\frac{5y}{x}=x^{6}"

This is in the form "\\frac{dy}{dx}+Py=Q"

"P=\\frac{-5}{x},Q=x^{6}"

Integrating factor, I.F="e^{\\int Pdx}"

I.f="e^{-5{\\int \\frac{1}{x}dx}}=e^{-5ln\\ x}=e^{ln\\ x^{-5}}=\\frac{1}{x^{5}}"

"y*I.f=\\int QI.f dx"

"y*\\frac{1}{x^{5}}=\\int x^{6}*\\frac{1}{x^{5}}dx"

"\\frac{y}{x^{5}}=\\int xdx"

"\\frac{y}{x^{5}}=\\frac{x^{2}}{2}+C"

"y=\\frac{x^{7}}{2}+Cx^{5}"


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