Find the particular solution of the differential equation satisfying the initial condition
Answers
Answer:
Step-by-step explanation:
How to solve the separable differential equation and find the particular solution satisfying the initial condition y(−4)=3 ?
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Calculus Applications of Definite Integrals Solving Separable Differential Equations
2 Answers
Answer:
General Solution:
(
4
y
+
13
)
1
2
−
2
x
=
C
1
Particular Solution:
(
4
y
+
13
)
1
2
−
2
x
=
13
Explanation:
From the given differential equation
y
'
(
x
)
=
√
4
y
(
x
)
+
13
take note, that
y
'
(
x
)
=
d
y
d
x
and
y
(
x
)
=
y
, therefore
d
y
d
x
=
√
4
y
+
13
divide both sides by
√
4
y
+
13
d
y
d
x
(
1
√
4
y
+
13
)
=
√
4
y
+
13
√
4
y
+
13
d
y
d
x
(
1
√
4
y
+
13
)
=
1
Multiply both sides by
d
x
d
x
⋅
d
y
d
x
(
1
√
4
y
+
13
)
=
d
x
⋅
1
d
x
⋅
d
y
d
x
(
1
√
4
y
+
13
)
=
d
x
⋅
1
d
y
√
4
y
+
13
=
d
x
transpose
d
x
to the left side
d
y
√
4
y
+
13
−
d
x
=
0
integrating on both sides we have the following results
∫
d
y
√
4
y
+
13
−
∫
d
x
=
∫
0
1
4
⋅
∫
(
4
y
+
13
)
−
1
2
⋅
4
⋅
d
y
−
∫
d
x
=
∫
0
1
4
⋅
(
4
y
+
13
)
−
1
2
+
1
(
1
−
1
2
)
−
x
=
C
0
1
2
⋅
(
4
y
+
13
)
1
2
−
x
=
C
0
(
4
y
+
13
)
1
2
−
2
x
=
2
⋅
C
0
(
4
y
+
13
)
1
2
−
2
x
=
C
1
General Solution
But
y
(
−
4
)
=
3
means when
x
=
−
4
,
y
=
3
We can now solve for
C
1
to solve for the particular solution
(
4
y
+
13
)
1
2
−
2
x
=
C
1
(
4
(
3
)
+
13
)
1
2
−
2
(
−
4
)
=
C
1
C
1
=
13
Therefore , our particular solution is
(
4
y
+
13
)
1
2
−
2
x
=
13
God bless....I hope the explanation is useful.