Differential Equations

The highest derivative is y⁽⁴⁾, so the order is 4.

The equation is not polynomial in the derivatives because it contains sin(y⁽³⁾).

The highest derivative is y′.

It appears to the first power in a polynomial equation.

The highest derivative is d²s/dt², so the order is 2.

That derivative has power 1.

The highest derivative is y″, so the order is 2.

cos(y′) makes the equation non-polynomial in derivatives.

The highest derivative is y″, of order 2.

It occurs to the first power.

The highest derivative is y‴, so the order is 3.

Its power is 3; all derivative terms are polynomial.

The highest derivative is y‴.

It occurs to the first power.

The highest derivative is y′.

It appears linearly.

The highest derivative is y″.

It is to the first power and the equation is polynomial in derivatives.

The highest derivative is y″.

It is to the first power; sin y contains no derivative.

The highest derivative is y″, so the order is 2.

sin(y′) means the equation is not polynomial in derivatives.

The highest derivative present is y″.

y′=eˣ and y″=eˣ.

Thus y″−y′=eˣ−eˣ=0.

y′=2x+2.

Substitution gives 2x+2−2x−2=0.

y′=−sin x.

Then y′+sin x=0.

y′=x/√(1+x²)=x/y.

Since y²=1+x², xy/(1+x²)=x/y.

y′=A.

Then xy′=Ax=y.

y′=sin x+x cos x, so xy′=x sin x+x²cos x.

Also √(x²−y²)=|x cos x|. The right side equals x sin x+x|x cos x|.

Differentiate: y+xy′=y′/y.

Rearrange: y²=(1−xy)y′, hence y′=y²/(1−xy).

Differentiate y−cos y=x: (1+sin y)y′=1.

Using x=y−cos y, the coefficient y sin y+cos y+x becomes y(1+sin y).

Differentiate: 1+y′=y′/(1+y²).

Rearranging gives y²y′=−(1+y²).

Differentiate y²=a²−x²: y′=−x/y.

Then x+yy′=x−x=0.

A general solution has one independent arbitrary constant for each order.

A particular solution has the constants fixed by conditions.

Use (1−cos x)/(1+cos x)=tan²(x/2)=sec²(x/2)−1.

Integrate to obtain y=2tan(x/2)−x+C.

Separate: dy/√(4−y²)=dx.

Integrate: sin⁻¹(y/2)=x+C.

dy/(1−y)=dx.

Integrate: −ln|1−y|=x+C.

d(tan x tan y)=sec²x tan y dx+tan x sec²y dy.

Therefore the differential is d(tan x tan y)=0.

dy/dx=(eˣ−e⁻ˣ)/(eˣ+e⁻ˣ)=d/dx[ln(eˣ+e⁻ˣ)].

dy/(1+y²)=(1+x²)dx.

Integrate both sides.

dy/(y ln y)=dx/x.

Integrating gives ln|ln y|=ln|x|+C.

For y≠0, y⁻⁵dy=−x⁻⁵dx.

Integrate and rearrange.

Integrate by parts with u=sin⁻¹x and dv=dx.

∫sin⁻¹x dx=x sin⁻¹x+√(1−x²).

Rearrange to (sec²y/tan y)dy=−eˣ/(1−eˣ)dx.

Integrate both logarithmic derivatives.

Factor the coefficient: (x+1)(x²+1).

The quotient is 1/[2(x+1)]+(3x−1)/[2(x²+1)].

Integrate and use y(0)=1.

1/[x(x²−1)]=−1/x+1/[2(x−1)]+1/[2(x+1)].

Integrate: y=(1/2)ln|x²−1|−ln|x|+C.

Use x=2,y=0 to fix C.

For real solutions, |a|≤1. A constant slope can be chosen as y′=cos⁻¹a.

Integrate and use y(0)=1.

dy/y=tan x dx.

ln|y|=−ln|cos x|+C.

Integrate: ∫eˣsin x dx=(eˣ/2)(sin x−cos x)+C.

Use (0,0) to obtain C=1/2.

Separate: dy/(y+2)=(1+2/x)dx.

Integrate: ln|y+2|=x+2ln|x|+C.

Use y+2=1 at x=1.

y y′=x, so y dy=x dx.

Integrate y²−x²=C; the point gives C=4.

y′=2(y+3)/(x+4).

Separate and integrate: ln|y+3|=2ln|x+4|+C.

The point gives y+3=(x+4)².

V=(4/3)πr³ is linear in time when dV/dt is constant.

r³ changes from 27 to 216 in 3 time units, so r³=27+63t.

P′=(r/100)P gives P=P₀e^{rt/100}.

At t=10, e^{10r/100}=2, hence r=10ln2.

P=1000e^{0.05(10)}.

N=N₀e^{kt}; e^{2k}=1.1, so k=ln(1.1)/2.

For doubling, e^{kt}=2.

e⁻ʸdy=eˣdx.

Integrate: −e⁻ʸ=eˣ+C.

Put y=vx, so y′=v+xv′.

Then x v′=(1−v)/(1+v), and ∫(1+v)/(1−v)dv=∫dx/x.

Set y=vx. Then v+xv′=1+v.

Thus v′=1/x.

With v=y/x, y′=(1+v)/(1−v).

Then x v′=(1+v²)/(1−v); integrate.

y′=(y²−x²)/(2xy). Set y=vx.

Then x v′=−(1+v²)/(2v), so ln(1+v²)=−ln|x|+C.

Set y=vx; then y′=v+xv′.

This gives x v′=1−2v², so separate dv/(1−2v²)=dx/x.

Put y=vx, so xdy−y dx=x²dv.

For x>0, √(x²+y²)=x√(1+v²); hence dv/√(1+v²)=dx/x.

Set v=y/x and dy/dx=v+xv′.

After cancellation, (tan v−1/v)dv=2dx/x.

Integrate both terms.

Set y=vx; then xy′−y=x²v′.

Separate dv/sin v=−dx/x.

Set v=y/x and dy/dx=v+xv′.

Separate to get (ln v−2)/(v(ln v−1))dv=−dx/x.

Integrate and simplify.

Put u=x/y, so dx=u dy+y du.

The dy terms combine to u+eᵘ, while the remaining term is y(1+eᵘ)du.

Recognise d[y(u+eᵘ)]=0.

Set v=y/x. Then x v′=−(1+v²)/(1+v).

Integrate: (1/2)ln(1+v²)+tan⁻¹v=−ln x+C.

Use (1,1).

With v=y/x, the equation gives x v′=−v(v+2).

Integration yields (v+2)/v=Cx²; the condition gives C=3.

Let v=y/x; the equation reduces to x v′=−sin²v.

Integrate csc²v dv=−dx/x, giving cot v=ln x+C.

Use v=π/4 at x=1.

Let v=y/x, so y′=v+xv′.

The equation reduces to x v′=−cosec v.

Integrate sin v dv=−dx/x and apply the condition.

Put v=y/x. Then x v′=v²/2.

Integrate −2/v=ln x+C. The initial value gives C=−1.

The right side depends on x/y.

Set v=x/y, equivalently x=vy.

In D, every coefficient is a homogeneous polynomial of degree 2.

The other options contain constants or terms of different total degrees.

The integrating factor is e^{2x}.

Integrate (ye^{2x})′=e^{2x}sin x.

The integrating factor is e^{3x}.

(ye^{3x})′=eˣ, so ye^{3x}=eˣ+C.

For x≠0 the integrating factor is x.

(xy)′=x³.

An integrating factor is sec x+tan x.

Integrate (y(sec x+tan x))′=tan x(sec x+tan x).

Divide by cos²x; the integrating factor is e^{tan x}.

With u=tan x, ∫u eᵘdu=eᵘ(u−1).

The integrating factor is e^{2x}.

Integrate (ye^{2x})′=x²log x·e^{2x}.

Divide through by xlog x; the coefficient of y is 1/(xlog x).

The integrating factor is log x, and (ylog x)′=2log x/x².

Divide by dx and use integrating factor 1+x².

Then [y(1+x²)]′=cot x.

Divide by x; the integrating factor for y′+(1/x+cot x)y=1 is x sin x.

(xy sin x)′=x sin x.

Treat x as a function of y: dx/dy−x=y.

The integrating factor is e⁻ʸ; integrate (xe⁻ʸ)′=ye⁻ʸ.

Divide by dy: dx/dy+x/y=y.

The integrating factor is y; (xy)′=y².

Write dx/dy−x/y=3y.

The integrating factor is 1/y; hence (x/y)′=3.

The integrating factor is sec²x.

(y sec²x)′=sin x sec²x=tan x sec x=(sec x)′.

Integrate to get y sec²x=sec x+C. The condition gives C=−2.

The left side is [y(1+x²)]′.

Integrate: y(1+x²)=tan⁻¹x+C; use y(1)=0.

The integrating factor is sin⁻³x.

(y csc³x)′=2cos x/sin²x, so y csc³x=−2csc x+C.

The condition gives C=2.

y′=x+y, so y′−y=x.

The integrating factor is e⁻ˣ; integrate (ye⁻ˣ)′=xe⁻ˣ.

Interpret the stated condition as y′=x+y−5.

Solve y′−y=x−5 with integrating factor e⁻ˣ.

Use y(0)=2.

Divide by x: y′−y/x=2x.

The integrating factor is exp(∫−1/x dx).

Divide by 1−y²: dx/dy+[y/(1−y²)]x=ay/(1−y²).

The integrating factor is exp(∫y/(1−y²)dy).

(i) highest derivative y″, power 1; (ii) y′ has highest order and power 3; (iii) highest derivative y⁽⁴⁾.

For (iii), sine of a derivative makes the equation non-polynomial in derivatives.

(i) Put u=xy; then xy″+2y′=u″. Since u=aeˣ+be⁻ˣ+x², we have u″−u+x²−2=0.

(ii) Differentiate twice; the terms cancel in y″−2y′+2y.

(iii) y″=6cos3x−9xsin3x.

(iv) Differentiate the implicit relation and use x²=2y²log y.

Differentiate the proposed relation implicitly.

Collect the dx and dy terms and factor their common terms.

The resulting ratio is equivalent to the stated differential equation.

Separate: dy/√(1−y²)=−dx/√(1−x²).

Integrate both inverse-sine derivatives.

Differentiate the ratio (x+y+1)/(1−x−y−2xy).

Setting its derivative to zero gives y′=−(y²+y+1)/(x²+x+1).

The left side is −d(cos x cos y).

Thus cos x cos y=C; use the given point.

Separate: dy/(1+y²)=−eˣdx/(1+e²ˣ).

Integrate to get tan⁻¹y=−tan⁻¹(eˣ)+C.

At x=0,y=1, C=π/2.

Put u=x/y, so dx=u dy+y du.

Cancel the common u y eᵘdy terms to get eᵘdu=dy.

Let s=x+y and t=x−y. Then dx+dy=ds and dx−dy=dt.

The equation becomes t ds=dt, so s=ln|t|+C.

At (0,−1), s=−1,t=1, hence C=−1.

Invert to get y′+y/√x=e⁻²√x/√x.

The integrating factor is e^{2√x}.

Then (ye^{2√x})′=1/√x.

The integrating factor is sin x.

(y sin x)′=4x; integrate and apply the condition.

Separate: dy/(2eʸ−1)=dx/(x+1). Let u=eʸ.

Integration gives ln|(2eʸ−1)/eʸ|=ln|x+1|+C.

Use x=0,y=0 to determine C.

Multiply by y: y dx−x dy=0.

This is equivalent to d(x/y)=0.

The integrating factor is μ(y)=e^{∫P₁dy}.

Then (xμ)′=Q₁μ. Integrate with respect to y.

Divide by dx: y′+y=−2xe⁻ˣ.

The integrating factor is eˣ, so (yeˣ)′=−2x.