Definition 1 (see [24].)For any z(w) ∈ C[I, R], we defined the derivative of Caputo-Fabrizio for nonsingular kernel as

Definition 2 (see [24].)The integral of Caputo-Fabrizio for nonsingular kernel type is given by

Definition 3 (see [25].)Let n < μ ≤ n + 1 and f be such that f⁽ⁿ⁾ ∈ H¹(c, d). Set α = μ − n. Then, α ∈ [0, 1] and we define

Lemma 4. For z(w) defined on [c, d] and μ ∈ [n, n + 1], for some n ∈ N₀, we have

Lemma 5. The solution of

Proof. Let z(w) be a solution to problem (6). Applying Caputo-Fabrizio integral on both sides and then using Lemma 4 and Definition 3, we have

Using boundary conditions z(c) = z(d) = 0, we have

Putting c₀, c₁ in (9), we get

Corollary 6. In view of 6, the solution of the considered problem (1) is given by

Further, for the existence and uniqueness of the solution of problem (1), we use some fixed point theorems. For this, we need to define an operator as N : C[I, R] → C[I, R] by

To proceed further, using Corollary (6) to convert the proposed problem (1) is to a fixed point problem as Nz(w) = z(w), where the operator N is given by (13). Therefore, Problem (1) has a solution if and only if the operator N has a fixed point, where λ(w) = f(w, z(w), λ(w)) and $$. We assume that

(H₁) There exist certain constant D_f > 0 and 0 < E_f < 1, such that

Theorem 7. Under the hypothesis (H₁), the mentioned problem (1) has a unique solution if

Proof. Suppose $$ we have

Repeating the above process, we get

Using (18) in (16), we have

Applying maximum on both sides, we have

Thus, operator N is a contraction; therefore, the operator N has a unique fixed point. Hence, the corresponding problem (1) has a unique solution.

Theorem 8 (see [26].)Let H ⊂ C[I, R] be a closed, convex nonempty subset of C[I, R]; then, there exist N₁, N₂ operators such that

• (1)

  N₁z₁ + N₂z₂ ∈ H for all z₁, z₂ ∈ H

• (2)

  N₁ is a contraction, and N₂ is compact and continuous

Then, there exist at least one solution z ∈ H such that N₁z + N₂z = z.

Theorem 9. If the hypothesis (H₂) is satisfied, then (1) has at least one solution if

Proof. Suppose we define two operators from (13) as

Let us define a set F = {z ∈ C[I, R]: ∥z∥≤r}, since f is continuous, so we show that the operator N₁ is contraction. For this $$ we have

Hence, N₁ is contraction. Next, to prove that the operator N₂ is compact and continuous, for this z(w) ∈ C[I, R], we have

Now, using (28) in (26) and then taking the maximum on both sides, we have

Which implies that

Therefore, N₂ is bounded. Next, let w₁ < w₂ in I, we have

Now, using (28) in (31), we have

Applying maximum on right-hand side of the above inequality, we take

Obviously, from (33), we see that w₁ → w₂; then, the right-hand side of (33) goes to zero, so ∣N₂z(w₂) − N₂z(w₁) | →0 as w₁ → w₂. Hence, the operator N₂ is continuous. Also, N(H) ⊂ H; therefore, the operator N₂ is compact, and by the Arzela-Ascoli theorem, the operator N has at least one fixed point. Therefore, the mentioned problem (1) has at least one solution.

Definition 10. The proposed problem (1) is Hyers-Ulam stable if at any ε > 0 for the given inequality

Further, the considered problem (1) will generalize Hyers-Ulam stable if there exists nondecreasing function ϕ : (c, d) → (0, ∞) such that

Remark 11. Let there exist a function ψ(w) which depends on z ∈ C[I, R] with ψ(c) = 0 and ψ(d) = 0 such that

Lemma 12. The solution of the given proposed problem

Proof. The solution of (39) can be acquired straightforward by using Lemma 5. Although from the solution, it is clear to become result (40) by using Remark 11.

Theorem 13. Under the Lemma 12, the solution of the proposed problem (1) is Hyers-Ulam stable and also generalized Hyers-Ulam stable if $$

Proof. Let z(w) ∈ C[I, R] be any solution of the considered problem (1) and $$ be a unique solution of the said problem; then, we take,

Using (40) and (18) in the above inequality, then taking maximum on both sides, we have

Hence, from the above inequality, we have

Therefore, the solution is Hyers-Ulam stable. Further, let

Example 14. Suppose, we take the boundary value problem of implicit type as

Clearly, c = 0, d = 1 and $$ is a continuous function for all x ∈ [0, 1]. Further, suppose that $$; then, we consider as

which implies that

Since from (48), one has D_f = 1/55, E_f = 1/55, and μ = 1/3. Further, also consider

Therefore, p_f = 1/35, q_f = 1/55, r_f = 1/55. and G_μ = 1/3, $$, c = 0, and d = 1. Then

Therefore, the conditions of Theorem 7 are satisfied. Thus, the problem (46) has a unique solution. Further, we need to satisfy some conditions of theorem (9).

Hence, the conditions of Theorem 9 also hold. Therefore, (46) has at least one solution. Furthermore, proceed to verify the stability results; we see that

Example 15. Take another boundary value problem of implicit type as

Clearly c = 0, d = 1 and $$ is a continuous function for all w ∈ [0, 1]. Further let $$, then consider, we have

Thus from (55), one has D_f = 1/45, E_f = 3/65, and μ = 3/7. And also consider we have

Therefore, the conditions of Theorem 7 are satisfied. Thus, the problem (53) has a unique solution. Further, we need to satisfy some conditions of Theorem (9), we have

Hence, the conditions of Theorem (9) also hold. Therefore, (53) has at least one solution. Furthermore, proceed to verify stability results; we see that

Hence, the solution of the mentioned problem (53) is Hyers-Ualm stable and consequently generalized Hyers-Ulam stable.