First Order and first-degree differential equation
A differential equation of the form
M+N dydx=0M+N dydx=0
or
M dx+N dy=0M dx+N dy=0
is called first-order and first-degree differential equation where both
MM
and
NN
are functions of x and y or constants.
Types of differential equation of first order and first degree:
(1) Separation of variables
(2) Homogeneous equation
(3) Equation reducible to homogeneous
(4) Exact equation
(5) Linear equation
(6) Reducible to linear equation.
Variable Separable:
If the differential equation
M dx+N dy=0M dx+N dy=0
can be expressed in the form
f(x) dx+ϕ(y) dyf(x) dx+ϕ(y) dy
- then it is called variables are separable.
Rule 1: If the differential equation is of the form
f(x) dx+f(y) dy=0f(x) dx+f(y) dy=0
ie function of xx with dxdx and function of yy with dydy, then only integrate we will find the solution
ie
f(x) dx+f(y) dy=0∫f(x) dx+∫f(y) dy=0F(x)+F(y)=cwhen,∫f(x) dx=F(x)
Question 1: Solve,
dydx+y2+y+1x2+x+1=0
Solution: Given equation is
dydx+y2+y+1x2+x+1=0⟹dydx=−y2+y+1x2+x+1⟹dyy2+y+1=−dxx2+x+1⟹dyy2+y+1+dxx2+x+1=0
Integrating both sides.
∫dyy2+y+1+∫dxx2+x+1=0⟹∫dy(y+12)2+(√32)2+∫dx(x+12)2+(√32)2=0⟹1√32tan−1(y+12√32)+1√32tan−1(x+12√32)=c⟹2√3tan−1(2y+1√3)+2√3tan−1(2x+1√3)=c
Rule 2: If any differential equation of the form
f(x) ϕ(y) dx+f(y) ϕ(y) dy=0
then divide the equation by
ϕ(x) ϕ(y)
. the equation of the form
f(x)ϕ(x) dx+f(y)ϕ(y) dy=0
ie function of x with dx and function of y with dy. Then only integrate we can find the solution.
Question 2: solve,
x√1−y2 dx+y√1−x2 dy=0
Solution: Given equation is
x√1−y2 dx+y√1−x2 dy=0
Dividing both sides by
√1−x2 √1−y2
X DX√1−x2 dx+y dy√1−y2=0⟹−12∫−2x dx√1−x2−12∫−2y dy√1−y2=0⟹−12 2√1−x2−12 2√1−y2=−c [Since f′(x) dx√f(x)=2√f(x)]⟹√1−x2+√1−y2=c
Question 3: Solve,
sec2x tany dx+sec2y tanx dy=0
Solution: Given equation is,
sec2x tany dx+sec2y tanx dy=0
Dividing both sides by
tanx .tany
we get,
sec2x dxtanx+sec2y dytany=0
Now integrating both sides,
∫sec2x dxtanx+∫sec2y dytany=0⟹ln(tanx)+ln(tany)=lnc [Since ∫f′(x)f(x) dx=ln(f(x))]⟹ln(tanx .tany)=lnc⟹tanx .tany=c
Rule 3: If the differential equation is of the form
dydx=f(ax+by+c)
then put
ax+by+c=z
and simplify. Finally integrating we will get the solution.
Question 4: Solve,
sin−1(dydx)=x+y
Solution: Given equation is
sin−1(dydx)=x+y⟹dydx=sin(x+y)
We put
x+y=z
Then
1+dydx=dzdx⟹dydx=dzdx−1
From (4.1) and (4.2) we get,
dzdx−1=sinz⟹dzdx=1+sinz⟹dz1+sinz=dx
Integrating both sides,
∫dz1+sinz=∫dx⟹∫1−sinz1−sin2z dz=∫dx⟹∫1−sinzcos2z dz=∫dx⟹∫sec2z dz ∫secz .tanz dz=∫dx⟹tanz−secz=x+c⟹tan(x+y)−sec(x+y)=x+c
Question 5: Solve,
dydx=(4x+y+1)2
Solution: Given equation is
dydx=(4x+y+1)2
we put
4x+y+1=z
then
4+dydx=dzdx⟹dydx=dzdx−4
From (5.1) and (5.2) we get
dzdx−4=z2⟹dzdx=4+z2⟹dz4+z2=dx
Integrating both sides we get,
12tan−1z2=x+c2⟹tan−1z2=2x+c⟹ z2=tan(2x+c)⟹4x+y+1=2tan(2x+c)
Question 6: Solve,
(ex+1)y dy=(y+1)ex dx
Solution: The given equation is
(ex+1)y dy=(y+1)ex dx⟹y dyy+1=ex dxex+1⟹(y+1−1) dyy+1=ex dxex+1⟹(1−1y+1) dy=d(ex+1)ex+1∴dy−dyy+1=d∗ex+1ex+1
Where the variables have been separated, so the general solution will be obtained by integrating on both sides and adding a constant.
∫dy−∫dyy+1=∫d(ex+1)ex+1y−ln(y+1)=ln(ex+1)+lnc [wherelnc is the integrating constant]y=ln(y+1)+ln(ex+1)+lncy=ln(y+1)(ex+1).cey=(y+1)(ex+1)c
Which is the required solution.
Question 7: Solve,
y(1+xy) dx+x(1−xy) dy=0
Solution: The given equation is,
y(1+xy) dx+x(1−xy) dy=0⟹y dx+xy2 dx+x dy−x2y dy=0⟹x dy+y dx+xy(y dx−x dy)=0⟹d(xy)+xy(y dx−x dy)=0⟹d(xy)(xy)2+dxx−dyy=0 [Dividing (xy)2 on both sides]
Here the variables have been separated, so the general solution will be obtained by integrating on both sides and adding a constant
∫d(xy)(xy)2+∫dxx−∫dyy=0⟹−1xy+lnx−lny=c⟹lnx−lny−1xy=c⟹lnxy−1xy=c
Which is the required solution.
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